It also helps to find coordinates of P and Q by adding/subtracting a vector AB, normalized to corresponding radii, from coordinates of A and B Oh, so just like the officials proposed it to be solved.
3:08 "No obvious way to do this" Uh, I think it should be immediately obvious that the shortest distance between two circles would be along a straight line through their centers.
It's a fairly obvious idea to think of, sure. To a lot of people anyway. But there's a difference between coming up with the idea, and coming up with a way to prove that the idea is right Sometimes things in math look pretty obvious, but actually turn out to be wrong, so mathematicians don't believe something just because it looks like it's true, even if it feels really really obvious. They still check whether they can actually prove it And I don't think the proof is obvious. It's not all that hard to come up with a proof, but it's also not so obvious that you can just take it for granted
This can be solved within 1 - 2 minutes. The distance between centers, that is the distance between (5,4) and (1,1) is simply sqrt((5-1)^2+(4-1)^2) = 5 Now subtract the two radii from 5, which are 2 and 1 from 5 We get 5-2-1=2
@@kimba381You also know that the right side of the circle equations is r². Thus the first circle has a radius of sqrt(4) = 2 and the second of sqrt(1) = 1, hence you have the distance between the perimeters of 5 - 2 - 1 = 2.
Yeah, but this wasn't an exam question, it was an interview question. You're there to have a conversation with the interviewers and explore the problem. You don't _want_ to solve it in two minutes. You want to use the opportunity to show your skills Finding the solution is easy. If you want to impress, you need to be able to prove that it's the right solution
@douglaswolfen7820 I would rather show my ability to solve a problem in a fast and efficient way. To show my skills, first I will solve it by the same way (that is in 1-2 minutes), then I will spend some time in proving the distance formula, sqrt(...^2+....2). And will spend some time in proving that the shortest distance is indeed lies on the centres of the two circles. Spending time can be in some related things, not in solving the problem itself. (My opinion). Thanks for sharing your thoughts.
You made this WAY too hard. 1. Find the distance between the two points and subtract the radii. 2. Find the equation on the line between the radii and determine where it intersects each circle.
I agree. In addition, the conclusion at the end of the video ('and that's the answer!') does not answer the original question, which is "what's the shortest distance".
I actually did this in my head. Once it was clear that the centers of the two circles had a difference of 3 and 4, the resulting hypoteneuse had to be 5, which is the distance between the two centers. Subtract the two radii (1 and 2), and you get 2. Figuring out the coordinates of P and Q followed the same logic, since the triangles are all similar (3,4,5; 3/5, 4/5, 1; 6/5, 8/5, 2); so the coordinates of Q are (1.8, 1.6) and the coordinates of P are ( 3.4, 2.8).
i have done such problems in an entrance exam mentally. they were not impressed and thought i must have cheated. I thought it easier to introduce v=x-1 and w=y-1. I’m not sure if it was quicker but it clarified things for me
And for part 2, an equivalent approach to the vector stuff is to draw vertical lines through P and Q, then work with smaller triangles that are similar to the 3-4-5.
Shortest path is the line segment connecting the two centers. Subtract the radius of both circles from that line segment to get the distance. Don't even need to plot anything.
Step 1: You have to consider the line between the centers of the circles, where one must be at (5,4) and the other at (1,1) given a simple visual inspection of the formulae provided. Step 2: Delta y and delta x between the two points are 4 and 3, respectively, so the length between the centers is simple since it represents a 3,4,5 right triangle, i.e., the total length is 5. Step 3: Immediately upon inspection again of the formulae you know that the radius of one circle is 1 and the other is 2 so you subtract the sum of these two radii from the total leading to a final answer of length = 2. Great question and great video SO A BIG THANKS AS ALWAYS! but I would skip the circle diagrams. Instead just look at it analytically this way as it saves a ton of time and avoids unnecessary complexity.
I'd take the equation for the line, and tried to find a px and py thats both on the line and the circke equation. the vector variant was proberly easier to calculate.
I did the same. Far easier to add the vectors, and I had not considered it. Also, when he said vectors, I was going to make it more difficult by changing the vector into polar coordinates and then back to cartesian coordinates. And I complain that my students make problems harder than they need be.
I did the first part in my head while looking at the equations. The centers are at (5, 4) and (1, 1) so the line between them is the hypotenuse of a 3-4-5 triangle, so the distance is 5. Subtract the radius of each circle and you get 5-2-1=2. (And *that's* the answer). This works because the shortest distance between two circles is on a line perpendicular to both circle, so it is on the line between the centers.
I love how you go into agonizing detail to show a the shortest distance between two points is a straight line (ugh), and then totally gloss over how to add vectors.
It's actually the inverse: * The shortest distance between the centers is a straight line, yup. * The shortest distance between two points on the edges is also a straight line, yup. * But unlike the centers, we have a choice of edge points. Which ones lead to the shortest straight line? The ones that lie along the straight line between the centers. That may be intuitively obvious, but the video actually spells out the algebra underlying it (minimizing X + a couple constants is equivalent to minimizing X).
He doesn't actually show the shortest distance between 2 points is a straight line. He uses that fact, however, to prove the minimum distance between 2 points on disjoint circles lie on such a line (since the radii are constant and therefore the distance between the points on the circles must also be minimum).
I surprised myself by figuring the first part out in my head. I tackled part 2 differently and couldn’t make it all the way through without writing some of it down. I make a linear equation connecting the centers of the circles, solved for y, and plugged the resulting expression into the two circle equations. Solving for x then yields four different x possibilities (two for each circle) and the closest two will be the desired points on the circumferences. Then you plug the x numbers back into the linear equation to solve for the two y coordinates. The vector solution here feels more intuitive, but I’m not sure I would have thought of that on my own.
Any approach that works, works. And this sounds like a totally valid approach, even if it was a bit more work Did you have a diagram to look at? Sketching it out on paper can sometimes really help you find the more intuitive solutions
I spent a couple of seconds thinking I was going to need trigonometry to figure out the coordinates of P and Q. Then my brain kind of short-circuited when I tried to figure out the details "But what's the angle? Oh wait… basic trig is about starting with an angle, and then using it to find the gradient. I don't actually know the angle, and I _already_ have the gradient. So I can skip that part"
@@douglaswolfen7820 I had a mental diagram going, and that was enough for part 1. I figured out the length of the line between the centers and subtracted off both radii to get the distance between the circumferences. But in part 2, I was tracking too many numbers and equations to have it stay in my head. That’s when the paper came out. 🙂
I am terrible at math. However, I watch just in case something gains access to that rock I call my 'math' brain... still blows me away. I am highly proficient in EVERYTHING but math. But damn committed to learning.❤
Start solving 10th grade problems of geometry or algebra by giving it some time, once you will start getting correct answers, your confidence will skyrocket which will increase your interest and, interest and peace of mind are the only things you need for problem solving.
Have you tried "3 Blue 1 Brown" videos? The topics are usually bit more advanced, but I think they do a much better job of explaining _why_ mathematicians would do the things they're doing Presh isn't always great at explaining what assumptions he's starting from, or why he's taking a particular approach
might I recommend you cut out a bunch of little squares of paper and treat these as number units.. It can really help people who don't grasp math as it is typically written down. but it will become visually obvious when you see, addition, multiplication, etc... it will also make it obvious why we call a number times itself 'squared'. If you do it with lego - cubed will become obvious too. It really helps non mathsy people grasp whats actually going on. I know it sounds quite childish.. but honestly its a REALLY good exercise for learning.
The shortest path between two circles lies on the line AB where A and B are the centerpoints of their respective circles. From there it's easy to find that line since we know the points are (1,1) and (5,4). (y = 3x+1/4). Substitute y with 3x+1/4 for the circle equations, solve, and plug the solutions that make sense (they are quadratics so there are 2 solutions each) back into 3x+1/4 to get x and y for the coordinates for P and Q. Then use the distance formula, in this case, 2. P = (17/5,14/5) and Q is (9/5,8/5), and the distance is 2.
1) find the distance between the circle's centers. Subtract the radius of each circle. that's the answer. 2) find the slope of the line from #1. solve a simultaneous system of the circle and that line. Do it again for the other circle. Those are the points. Plenty of algebra but straight-forward
A lot of people are missing the point. It's visually "obvious" that the shortest distance between the two circles should lie on the line connecting their centers. But *why*? The insight that the solution is obtained by translating one minimization problem to another is the satisfying one. Your response to this problem may indicate whether you are an engineer or a mathematician at heart.
I'm a physicist, and I find the solution WAY WAY too long and complicated. He ends up making the assumption, anyway, that the shortest distance between that complicated path is a straight line, without proving it. Might as well look at it and say "hell, just connect the center of the circles, it will give the answer"
Depends. Do you want to learn more about constructing a mathematical proof? Figuring out the answer in 5 seconds is good, but proving it takes more work. I think that's what the video is about
@@douglaswolfen7820 i did this exact logic in my head in like 15 seconds ( skipping the whole random PQ placement as its just obvious it needs to be a straight line)
@@ollie1505 ah yes, the old "it's just obvious" line of argument. Of course it's obvious. But do you know how to prove it? (Maybe you do. But some people don't, and some people would learn something from watching it proved)
I was actually able to solve this in my head when I realized… (spoiler) …that the centers of the circles are on a 345 triangle. Definitely one of those problems that looks harder than it is at first glance.
But that's how you would know the distance between the centers a 5, you would need to use a distance formula and points and stuff to find the distance of the soace between the 2 circles because that's the actual question.
This is a way more complicated solution that I had. Here’s what I did: 1) Take the circles center points and find the slope. This ends up being 3/4. 2) This slope created a special right triangle of side lengths of 3, 4 and a hypotenuse of 5. 3) Create a triangle scaled 1/5 and 2/5 the size. These will have a hypotenuse equal to the radii. 4) Accordingly add/subtract the x and y components of the triangles from the center points to get the answer.
This isn't even a circle problem. It's finding the distance between two desired points on a line between two given points, knowing the distance from the given points to their nearest desired point. This is just subtraction on a hypotenuse.
I guess the lenghty reasoning to solve Q1 is to demonstrate that the shortest distance is on a straight line between centers, which is intuitively obvious.
This is so easy I did this in my head in 30 seconds, 5-1=4, 4-1=3 -> special right triangle 3,4,5 -> 5-2-1=2 -> answer is 2. I'm not a genius but sometimes problems on this channel are easy, and sometimes are really hard. This was the easy one with a misleading "Oxford" caption.
I used this: -The two closest points always lie on a line that intersects both radii. -If the two closest points lie on both the line and the circles, then the points must be solutions of both. -Find the equation of the line using the two radii. y = (3/4)x + 1/4 -Substitute y in each circle’s equation for the line’s equation and solve for x and y. When solving for either variable, choose the point of the two solutions closest to the goal, the two points that would be closest to each other on the circles. For example, the first circle, (x-1)^2 + ((3/4)x + 1/4 - 1)^2 = 1 yields x = 0.2 and 1.8. I would choose 1.8 since I know the centers of the two circles, (1, 1,) and (5, 4), and would choose the most sensible points to aim for the shortest distance. -The two points would be (1.8, 1.6) and (3.4, 2.8), having a distance of 2. The key to this question was remembering the property of the closest distance between two circles being on the same line as a line through both diameters. :)
I like how you can use pythagerous for this. So process is. 1) Draw two circles on a graph at the stated centres and radii. (x-h)sq +(y-k)sq=rsq centre=(h,k) radius=(r) Circle 1 is origin 5,4 radius 2 Circle 2 is origin 1,1 radius 1 2)Draw a straight line point to point from the centre one circle to the other. Because this is the shortest point between PQ where we know the fixed lengths of both R. 3)Find the length of the line by using pythagerous theorum. 4)Subtract the radii of each circle from the line length. 5)This is the shortest point between the circumfrence of both circles. 6) Lables H= Hypotenuse, X= horizontal, Y=vetical, rSubscript1= first circle (larger one), rSubscript2=second circle (smaller circle) P= co-ordinate large circle Q= co-ordinate small circle A= Centre of Large Circle B= Centre of small Circle 7) Vectors which I am not going to simplify as it doesnt lend itself to long writing. Both are going to use point B as the vector origin. Vector BA=(4,3) [(BQ)/(BA)=1/5] Q= B+[1/5 of the vector BA] Q= 1,1+[1/5 of (4,3)] Q= 1.8,1.6 Vector BA=(4,3) [BP/BA)=3/5] Note H minus Rsubscript 1 makes point P the location on the hypotenues length from vector start point B P= B+[3/5 of vector BA] P=1,1 +[3/5 of 4,3] P=3.4, 2,8 Comment added for my own working through the process as a learning tool.
The coordinates of the points are so much easier to find. You can use basic trigonometry. 1. Find the angle of the line ab at point b. You can do this using the right angle triangle shown, with side lengths 4 and 5. 2. Use the angle of the line at point b and the distance between points b and p to create triangle with side lengths that are equal to the coordinates of point p. 3. Repeat for the other point
The shortest distance lye on line joining centers.Line joining centers is of length √4^2 + 3^2 =5. Shortest distance is 5 - (sum of radii of circles) = 5 -3=2.By forming and solving quadratic equations we get coordinates of P and Q are:(17/5 , 14/5) and (9/5, 8/5,)
gonna start by saying thanks for this problem! for the coordinates of the points q and p: first i used y=mx+b to get the equation of that intersecting line (slope can be found using the two center points of the circle) then set the line equation equal to the equation of the smaller circle to figure out the values of the 2 intersecting x points (and selected the greater of the 2) then back tracked to find the two intersecting y points of that x value along the equation of the same circle used from there the same logic could be applied to find out the values for coordinate p although cumbersome, it was enjoyable to relearn a bunch of old math concepts. Thanks! 😁
The shortest distance between circles is on the line through both centers which crosses the circles at a right angle. It is the distance between the centers minus the radii of both circles. From the formulas we know the origins (5,4) and (1,1) and the radii √4 (2) and √1 (1). The offset between the origins is (4,3) like the sides of the known 3-4-5 triangle so the distance must be the hypotenusa 5. Subtracting the radii 2 and 1 gives 2 as the distance between the circles.
Extremely easy. find distance of centres using distance formula, then subtract sum of radii from that distance c1=(5,4), r1=2 units; c2=(1,1), r2=1 unit d=root(((5-1)^2)+((4-1)^2)) =5 units r1+r2=3 units shortest distance=5-3=2 units.
The distance between the centers = sqrt((5-1)^2 + (4-1)^2) = 5. Since the radius of the first circle is sqrt(4) = 2 and the radius of the second circle is sqrt(1) = 1, the circle do not intersect. Hence, the closet two points lie on the line segment joining the two centers where that line segment intersects the two circles. That distance is, clearly, 5 - (2 + 1) = 2.
Yes, I got same: circle formula, so: Circle A is of radius 2 (square root of 4) and centered at (5,4). Circle B is of radius 1 (square root of 1) and centered at (1,1). Shortest distance D between the circles is distance from P to Q, which is distance between the centers of the two circles, less the radius of each D=((5-1)^2+(4-1)^2)^.5-2-1 D=(25)^.5-2-1 D=5-2-1 D=2 Draw line L through the radius of both circles, it also passes through points P and Q. L: y=Sx+C 4=5S+C 1=S+C 3=4S S=3/4 Slope of line is S=3/4 right triangle, 3*3+4*4=5*5, so sides 3, 4, 5 So, for distance 5 along line PQ, x goes up by 4, y goes up by 3, so, for distance 1 along line PQ, x goes up by 4/5, y goes up by 3/5, Q is distance radius of B (1) away from center of B (1,1) P is distance radius of B + D = 3 away from center of B (1,1) Q: (1+4/5,1+3/5) P: (1+3*4/5,1+3*3/5) Q: (1.8,1.6) P: (3.4,2.8)
For the line between the circles. Connect the circle centers with a line. Determine the length of that line. Subtract from that line the radius of both circles. The result is the length of the shortest line connecting the circles. In this case the line is the hypotenuse of a 3x4x5 triangle. 5 minus 1 minus 2 is 2. The line connecting the circles is length 2.
Centres (5,4) & (1,1). Distance between is 5. (Trivial 345 triangle). Rad 1, rad 2. Shortest distance between circles is 5-2-1=2. Done. No Pythag beyond the 345 triangle. No quadratics. Just basic geometry. 2:52
Haven't watched the video yet, but my first inclination is actually do this in reverse order: The shortest distance between two (non-overlapping) circles will always be along the line connecting their two center points, so find the equation for that line, then use the intersection between that line and each circle (two equations in two variables) to find the coordinates for each of P and Q, then use the distance formula to find out the distance between them. Though I suppose if you didn't need to know P and Q, you could find the distance by just calculating the distance between the centers, and then subtracting the radii of each circle... (might want to do that first anyway just to know that they're actually not overlapping (if the result is negative, they're overlapping, and the answer to the first part is 0))
Circle A has a radius of 2 and circle B has a radius of 1. Circle A’s center is at (5,4) and circle B’s is at (1,1). The distance between these two points is sqrt((4-1)^2 + (5-1)^2) = 5 units. Subtract both radii from this value and you get 2.
I just found the slope between the two centers of the circle, (1,1) and (5,4), using slope intercept form to get 4-1=m(5-1), 3=4m the slope is 3/4, plotting 3/4x we are not at the center, since plugging in 1 would get us 3/4, i add 1/4 to the line, ending up with 3/4x+1/4. Using this we can figure out the edges of the circle by substituting the y in each circle with the equation of line. Set each equation equal to 0, after solving for each solution, you get (0.2,0.4) and (1.8,1.6) for the smaller circle and (3.4,2.8) and (6.6,5.2). We can get rid of each extremes, as (1.8,1.6) and (3.4,2.8) are the closer points, the line connecting them is diagonal, which you could say is a hypotenuse of a triangle, subtracting the coordinates from each, you get a height of 1.6 and a length of 1.2, pythagorean theorem says a^2+b^2=c^2 or the square root of a^2+b^2=c, plugging in 1.6 and 1.2, you get 2.56+1.44, and that equals 4, and then you take the square root, which equals 2. The shortest distance between the two circles is 2.
I completely overcomplicated this by missing the fact that the vector between their centers is composed of the legs of a 3-4-5 triangle. Ended up finding P and Q the hard way first with the angle of this vector and the x-axis (transformed from polar coordinates) then the distance between P and Q. Got the correct answer at least!
If we can assume the shortest path is a straight line (which is a well known fact in flat space), it's easy. Draw a straight line from A to Q and call the intersection with the first circle P'. Note that AP and AP' are radii so they have the same length. If also QP and QP' have the same length, the triangle APQ is degenerate and P=P'. Otherwise, QP must be longer than QP' and we can shorten it by replacing P with P'. So If we assume QP is the shortest path, P must already be in the line AQ. Likewise, Q must be on the line BP. So AP, PQ, QB form a straight line and thus P,Q are on AB. The rest is simple calculation, √((4-1)² + (5-1)²) - √4 - √1 = 2.
The shortest distance will obviously be a straight line from a to b, i did a²+b²=c² and got PQ=2 For the coordinates i used trigonometry With θ and β being angels of the same triangle in Q1 So it will be Q ⇒ (1+cosθ,1+sinθ) ⇒(1+⅘,1+⅗)⇒(1.8 , 1.6) P⇒(5-2sinβ,4-2cosβ) ⇒(5-2.⅘ ,4-2.⅗) ⇒(3.4 , 2.8) Using similarity of the triangles
I'm definitely not the sharpest tool in the shed. But I still think that the approach you took to solve the first part of the problem is unnecessarily complicated.
Yes, there is no need to go into the optimization digression. APQB is a quadrilateral, so AB is smaller than the sum of the other three sides. But still an awesome video.
I used the slope of the line through their midpoints a tiny touch of trig to add/subtract to/from the coords of the midpoints. It was made trivial by the fact that it was a 3,4,5 triangle. Your way was a little cumbersome.
I got the answer (2) in my head in about four seconds. The centerpoints of the circles make a 3-4-5 right triangle, the radii are 1 and 2, so 5-1-2 = 2.
for the second question, I made a function for the hypotenuse (the line of the shortest distance between the two centers of the circles) and just used abit of trig to get the x values of q and p
If the circles are not intersecting, then those two points lies on the line that goes through their centres. We can create an equation for such a line and that gives us two points for each circle. We then pick the points that are nearest to each other. These two points also answer the first question, because their distance is also the shortest distance between those two circles, so you can solve second question and get answer to the first, or you can solve the first by calculation of distance of their centers and then subtracting the radius.
I solved this in like 2 minutes (toke the distance from center of circle A to center of circle B (hypotenuse of a right angle triangle) and substracted radius a ad radius B.... (didn't bother with part 2, no fun) I then watched the video and wondered why on earth the solution was strating out so complicated... honestly dropped of before the end The best way to teach math is to keep it simple
The shortest distance D is the length of the part P of the line L connecting the centers where P is measured from the intersections btw. L with A and B respectively. The length of L is the distance of the centers of CA = (5,4) and CB = (1,1) which is sqrt(4^2+3^2) = 5. The two radius RA=sqrt(4)=2 and RB=sqrt(1)=1 at both ends of L are not part of P so they must be subtracted from lengt od L. So length of P = D = 5-2-1 = 2. Takes no 10 seconds to do this in head.
It’s on a graph just create a right angle triangle and you see it has sides of 4 and 3 between the 2 centres with the hypothenuse being the min distance of 5.
Part 1: Those equations make it very straightforward. The first circle has center (5,4) and the second (1,1). The distance between those two points is 5. The radii of the two circles are 2 and 1 respectively, so that means the shortest distance between two points on the two circles is 5-2-1=2.
Part 2: To go from the center of A to B, you have to follow a line that goes 4 left and 3 down. If you follow 2/5th of that line, you land on point P, which is (5-4*2/5,4-3*2/5) = (3 + 2/5, 2 + 4/5). If you follow 4/5th of that same line, then you land on point Q, which is (5-4*4/5,4-3*4/5) = (1 + 4/5, 1 + 3/5).
It took 4 minutes to explain that the distance between two points is 5, and the radii around those points are total length 3. The second solution was interesting; I used trig ratios, but I like how it was done here.
What an ungainly solution to a fairly simple problem? Part A: Line between centres of circles must give the shortest distance because radii cut at 90 degrees. Shortest is simply pythagoras - total radii Part B is just similar triangles
15 seconds to solve. 1) Center of the first circle (x; y = 5; 4) Center of the second circle (x; y = 1; 1) 2) Distance between centers: √((5-1)^2 + (4-1)^2) = √(4^2 + 3^2) = √(16 + 9) = √25 = 5 3) Subtract the radii of the circles from the resulting distance, and get the shortest distance: 5 - 2 - 1 = 2 Answer: shortest distance is 2
How I solved part 1: Shortest distance between 2 points is a straight line, so points P and Q will be along Segment AB. A line down from point A and a line right from point B will intersect at 5,1 and form a right angle. Segment AB then forms the hypotenuse and makes this a right triangle whose sides are 3 and 4...this is a 3-4-5 right triangle. Segment AB is 5 units long. Circle A's radius is sqrt(4), so it's 2. Circle B's radius is sqrt(1), so it's 1. 5 - 2 - 1 = 2, so Segment PQ will be 2 units.
As soon as I saw the problem I instantly though, "lmao we have the distance formula for a reason, just subtract the radius" like idk why this is such a "hard" problem.
You can figure the centre and radius out by the equation itself and then calculate the distance from the centres then subtract the sum of both radii That'll give you the shortest distance
If the circles are plotted corrrectly, the distance from A to B is the length of the hypotenuse of a 3,4,5 triangle, so that distance is 5. Now we need only subtract the length of the two radii to get the distance between the circles: 5 - 2 - 1 = 2.
The distance between 2 circles (or any shapes) depends on where they are placed relative to each other! If you place them 1cm apart, then that's the distance between them. If you place them touching, then the distance is zero. So zero is the shortest distance between 2 circles...
A circle radius 2 at 5,4, and radius 1 at 1,1. Translate the 1,1 to 0, and the centre at that first circle is at 4,3. So, a pythagorean triple. Duh. Distance between the centres is 5, subtract the two radii (1 and 2), and we are left with a distance of 2. Let's watch the video and see how that goes. --edit-- oh, there's a second part. Looking at the line through the centers, we have that pythgorean triad triangle again inside each circle, the hypotenuse being the radius. So the x coordinates will be 1+4/5, 1+3/5 = (1.8, 1.6), and the coordiates of the second will be 5-2*4/5, 4-2*3/5, or (3.4, 3.8).
Lots of people more interested in saying they know how to do this, and uh way easier akshully, than people recognizing the point he's trying to make (how to think through a math problem). Its a lot less about bragging on right answers and more about teaching people how to think. This was a GREAT video for explaining to my 6th grader how the lesson is knowing how to combine simple math facts to think through a math problem.
The use of the expression "Daunting Problem" had me spending far too long thinking I had misunderstood the question, but no! it really was a straight line between the centres. Once I realised that I was thinking too hard it was easy and so was the rest of it. Is this an indictment of the university system or does "Oxford admissions" refer to the McDonalds in Oxford?
I only watched up to 2m20s. The 2 centres are corners of a triangle and we know the vertical and horizontal lengths so we know the diagonal length ie distance between centres. Then subtract the radius of each circle
The whole line drawing and point moving along the circle was harder to do than the whole math on this one. ;) Got the first part on looking at the question and the second one easily after. That said, i got a masters degree in maths and so this SHOULD be easy. 😅
I don't understand how it isn't obvious that the shortest distance lies on a line cutting through the middle of both circles. If the circles were on top of each other, vertical rather than on a diagnal, then it would be SUPER obvious because any other place on the circle would be further away, because the points on the line would be the highest or lowest point on each circle. Sadly, it gets more complocated trying to explain it than just looking at it. And calculating PQ is pretty easy after that. This reminds me of a problem in highschool Algebra where the teacher drew an isosceles(?) triangle and labeled the long side as 7 and the two short sides as 3. She asked us to give her the internal angles. One look, and I determined it was impossible because 3 + 3 = 6, which is less than 7, so this is an imoossible triangle. I got it wrong for not showing my work using trigonometry. Sometimes, logic is best. 😅
It's 2 am and I saw this ... Solved it in under a min (yea) Every circle equation is in the format (x-h)^2 + (y-k)^2 =r^2 Where the centre of the circle is (h,k) and radius is r After finding centres of both circles use distance formula to find distance √{(x-a)^2 + (y-b)^2} where (a,b) is centre of other circle.. now subtract sum of radiii of both circles from the distance calculated above and tada... All this is taught to 15 year olds in India who are preparing for IIT (a super difficult exam and I am preparing for it)
I have to admit I easily figured it out once the circles were shown on the graph, but when he referred to equations for circles, I had no idea what he was talking about until he showed it. In hindsight, I probably did learn it in school, but it did not stick with me at all.
The shortest distance between two disjoint circles always lies on the line between their centres.
That's what I thought, then you take the radii off that distance.
It also helps to find coordinates of P and Q by adding/subtracting a vector AB, normalized to corresponding radii, from coordinates of A and B
Oh, so just like the officials proposed it to be solved.
Yes... that is the rule we all memorise , but thanks to this video i understand now where this rule comes from
I almost got this one, so it wasn't that incredible.
Only if the two circles have no intersection together.
3:08 "No obvious way to do this"
Uh, I think it should be immediately obvious that the shortest distance between two circles would be along a straight line through their centers.
It's not obvious. Maybe intuitive, but not obvious.
took me 3 seconds to realize this
It's a fairly obvious idea to think of, sure. To a lot of people anyway. But there's a difference between coming up with the idea, and coming up with a way to prove that the idea is right
Sometimes things in math look pretty obvious, but actually turn out to be wrong, so mathematicians don't believe something just because it looks like it's true, even if it feels really really obvious. They still check whether they can actually prove it
And I don't think the proof is obvious. It's not all that hard to come up with a proof, but it's also not so obvious that you can just take it for granted
@@douglaswolfen7820 Thats a very nice thought! 😄😄
It's pretty obvious honestly
This can be solved within 1 - 2 minutes.
The distance between centers, that is the distance between (5,4) and (1,1) is simply sqrt((5-1)^2+(4-1)^2) = 5
Now subtract the two radii from 5, which are 2 and 1 from 5
We get 5-2-1=2
That was the easy bit tho.
@@kimba381The coord of points P and Q is far easier if you use the interception theorem
@@kimba381You also know that the right side of the circle equations is r². Thus the first circle has a radius of sqrt(4) = 2 and the second of sqrt(1) = 1, hence you have the distance between the perimeters of 5 - 2 - 1 = 2.
Yeah, but this wasn't an exam question, it was an interview question. You're there to have a conversation with the interviewers and explore the problem. You don't _want_ to solve it in two minutes. You want to use the opportunity to show your skills
Finding the solution is easy. If you want to impress, you need to be able to prove that it's the right solution
@douglaswolfen7820 I would rather show my ability to solve a problem in a fast and efficient way. To show my skills, first I will solve it by the same way (that is in 1-2 minutes), then I will spend some time in proving the distance formula, sqrt(...^2+....2). And will spend some time in proving that the shortest distance is indeed lies on the centres of the two circles. Spending time can be in some related things, not in solving the problem itself. (My opinion). Thanks for sharing your thoughts.
You made this WAY too hard. 1. Find the distance between the two points and subtract the radii. 2. Find the equation on the line between the radii and determine where it intersects each circle.
Using parametric formula of the segment AB is easier than finding the intersection of a line and a circle.
Oh
Its also NOT CLEAR fro the wuestion ifnone cirlce isninside thebother or.not right?
Because he needs to make the video 8 minutes long.😉
I agree. In addition, the conclusion at the end of the video ('and that's the answer!') does not answer the original question, which is "what's the shortest distance".
I actually did this in my head. Once it was clear that the centers of the two circles had a difference of 3 and 4, the resulting hypoteneuse had to be 5, which is the distance between the two centers. Subtract the two radii (1 and 2), and you get 2. Figuring out the coordinates of P and Q followed the same logic, since the triangles are all similar (3,4,5; 3/5, 4/5, 1; 6/5, 8/5, 2); so the coordinates of Q are (1.8, 1.6) and the coordinates of P are ( 3.4, 2.8).
Absolutely, correct approach, head math only.
i have done such problems in an entrance exam mentally. they were not impressed and thought i must have cheated.
I thought it easier to introduce v=x-1 and w=y-1. I’m not sure if it was quicker but it clarified things for me
Yeah, same here. Used the same logic. Doesn't take long, quite straight forward to be honest.
I did the same!
Nice!
Also AB is the hypotenuse of a 3-4-5 right triangle. We know the two sides from the graph are length 3 and 4, so AB is 5.
And for part 2, an equivalent approach to the vector stuff is to draw vertical lines through P and Q, then work with smaller triangles that are similar to the 3-4-5.
Shortest path is the line segment connecting the two centers. Subtract the radius of both circles from that line segment to get the distance. Don't even need to plot anything.
Step 1: You have to consider the line between the centers of the circles, where one must be at (5,4) and the other at (1,1) given a simple visual inspection of the formulae provided.
Step 2: Delta y and delta x between the two points are 4 and 3, respectively, so the length between the centers is simple since it represents a 3,4,5 right triangle, i.e., the total length is 5.
Step 3: Immediately upon inspection again of the formulae you know that the radius of one circle is 1 and the other is 2 so you subtract the sum of these two radii from the total leading to a final answer of length = 2.
Great question and great video SO A BIG THANKS AS ALWAYS! but I would skip the circle diagrams. Instead just look at it analytically this way as it saves a ton of time and avoids unnecessary complexity.
I'd take the equation for the line, and tried to find a px and py thats both on the line and the circke equation. the vector variant was proberly easier to calculate.
I did the same. Far easier to add the vectors, and I had not considered it. Also, when he said vectors, I was going to make it more difficult by changing the vector into polar coordinates and then back to cartesian coordinates.
And I complain that my students make problems harder than they need be.
This is great to refresh my memory on vectors and basic algebra for that matter I have forgotten a lot so I really like your channel thank you
Shift the axis by (1,1) so the small circle is at the origin, then find the minimum of the larger circle.
What exactly is the "minimum of the larger circle".
🤨
Minimum of the magnitude makes sense
You would prob need some kind of differential equation involving the magnitude of a parametric
I did the first part in my head while looking at the equations. The centers are at (5, 4) and (1, 1) so the line between them is the hypotenuse of a 3-4-5 triangle, so the distance is 5. Subtract the radius of each circle and you get 5-2-1=2. (And *that's* the answer).
This works because the shortest distance between two circles is on a line perpendicular to both circle, so it is on the line between the centers.
I agree with other comments that the answer was easy and can be worked out in our heads. However, I respect your approach as it helps beginners.
I love how you go into agonizing detail to show a the shortest distance between two points is a straight line (ugh), and then totally gloss over how to add vectors.
It's actually the inverse:
* The shortest distance between the centers is a straight line, yup.
* The shortest distance between two points on the edges is also a straight line, yup.
* But unlike the centers, we have a choice of edge points. Which ones lead to the shortest straight line? The ones that lie along the straight line between the centers. That may be intuitively obvious, but the video actually spells out the algebra underlying it (minimizing X + a couple constants is equivalent to minimizing X).
He doesn't actually show the shortest distance between 2 points is a straight line. He uses that fact, however, to prove the minimum distance between 2 points on disjoint circles lie on such a line (since the radii are constant and therefore the distance between the points on the circles must also be minimum).
That's not what he shows. Jesus, you people.
@@markkennedy9767thank you. You seem to be one of the few here who understands his proof. It's quite good.
It's hilarious that none of these 'math geniuses' answering your comment have managed to understand your point 😂
I surprised myself by figuring the first part out in my head. I tackled part 2 differently and couldn’t make it all the way through without writing some of it down.
I make a linear equation connecting the centers of the circles, solved for y, and plugged the resulting expression into the two circle equations. Solving for x then yields four different x possibilities (two for each circle) and the closest two will be the desired points on the circumferences. Then you plug the x numbers back into the linear equation to solve for the two y coordinates.
The vector solution here feels more intuitive, but I’m not sure I would have thought of that on my own.
Any approach that works, works. And this sounds like a totally valid approach, even if it was a bit more work
Did you have a diagram to look at? Sketching it out on paper can sometimes really help you find the more intuitive solutions
I spent a couple of seconds thinking I was going to need trigonometry to figure out the coordinates of P and Q. Then my brain kind of short-circuited when I tried to figure out the details
"But what's the angle? Oh wait… basic trig is about starting with an angle, and then using it to find the gradient. I don't actually know the angle, and I _already_ have the gradient. So I can skip that part"
@@douglaswolfen7820 I had a mental diagram going, and that was enough for part 1. I figured out the length of the line between the centers and subtracted off both radii to get the distance between the circumferences. But in part 2, I was tracking too many numbers and equations to have it stay in my head. That’s when the paper came out. 🙂
I am terrible at math. However, I watch just in case something gains access to that rock I call my 'math' brain... still blows me away. I am highly proficient in EVERYTHING but math. But damn committed to learning.❤
Start solving 10th grade problems of geometry or algebra by giving it some time, once you will start getting correct answers, your confidence will skyrocket which will increase your interest and, interest and peace of mind are the only things you need for problem solving.
Have you tried "3 Blue 1 Brown" videos? The topics are usually bit more advanced, but I think they do a much better job of explaining _why_ mathematicians would do the things they're doing
Presh isn't always great at explaining what assumptions he's starting from, or why he's taking a particular approach
Me too. Its fun being baffled though and knowing someone understands it.
might I recommend you cut out a bunch of little squares of paper and treat these as number units..
It can really help people who don't grasp math as it is typically written down.
but it will become visually obvious when you see, addition, multiplication, etc...
it will also make it obvious why we call a number times itself 'squared'.
If you do it with lego - cubed will become obvious too.
It really helps non mathsy people grasp whats actually going on.
I know it sounds quite childish.. but honestly its a REALLY good exercise for learning.
Keep it up!! And one day you will find you are not so terrible at math anymore
How good to know how incredibly useful this piece of information is going to be in every Uni student’s future life! Playing pool, for example …..
The shortest path between two circles lies on the line AB where A and B are the centerpoints of their respective circles.
From there it's easy to find that line since we know the points are (1,1) and (5,4). (y = 3x+1/4). Substitute y with 3x+1/4 for the circle equations, solve, and plug the solutions that make sense (they are quadratics so there are 2 solutions each) back into 3x+1/4 to get x and y for the coordinates for P and Q. Then use the distance formula, in this case, 2.
P = (17/5,14/5) and Q is (9/5,8/5), and the distance is 2.
And what an incredible explanation.
1) find the distance between the circle's centers. Subtract the radius of each circle. that's the answer.
2) find the slope of the line from #1. solve a simultaneous system of the circle and that line. Do it again for the other circle. Those are the points.
Plenty of algebra but straight-forward
Nice! I found the equation o0f the line and then the intersection of the curves. The vector solution is elegant.
Straight to the point. Good.
A lot of people are missing the point. It's visually "obvious" that the shortest distance between the two circles should lie on the line connecting their centers. But *why*? The insight that the solution is obtained by translating one minimization problem to another is the satisfying one. Your response to this problem may indicate whether you are an engineer or a mathematician at heart.
I'm a physicist, and I find the solution WAY WAY too long and complicated. He ends up making the assumption, anyway, that the shortest distance between that complicated path is a straight line, without proving it. Might as well look at it and say "hell, just connect the center of the circles, it will give the answer"
Does this even deserve a video? i just read the question and its done like literally in my head within like 5 secs.
Depends. Do you want to learn more about constructing a mathematical proof? Figuring out the answer in 5 seconds is good, but proving it takes more work. I think that's what the video is about
@@douglaswolfen7820 i did this exact logic in my head in like 15 seconds ( skipping the whole random PQ placement as its just obvious it needs to be a straight line)
@@ollie1505 ah yes, the old "it's just obvious" line of argument. Of course it's obvious. But do you know how to prove it?
(Maybe you do. But some people don't, and some people would learn something from watching it proved)
I was actually able to solve this in my head when I realized… (spoiler)
…that the centers of the circles are on a 345 triangle. Definitely one of those problems that looks harder than it is at first glance.
But that's how you would know the distance between the centers a 5, you would need to use a distance formula and points and stuff to find the distance of the soace between the 2 circles because that's the actual question.
@@maxhagenauer24 no need for a distance formula when known catheti are 3 and 4.
@@feedbackzaloop Catheti? What is that?
@@maxhagenauer24 plural of cathetus - sides of a right triangle, adjecent to a 90 degree corner.
Same here.
This is a way more complicated solution that I had. Here’s what I did:
1)
Take the circles center points and find the slope. This ends up being 3/4.
2)
This slope created a special right triangle of side lengths of 3, 4 and a hypotenuse of 5.
3)
Create a triangle scaled 1/5 and 2/5 the size. These will have a hypotenuse equal to the radii.
4)
Accordingly add/subtract the x and y components of the triangles from the center points to get the answer.
This isn't even a circle problem. It's finding the distance between two desired points on a line between two given points, knowing the distance from the given points to their nearest desired point. This is just subtraction on a hypotenuse.
I guess the lenghty reasoning to solve Q1 is to demonstrate that the shortest distance is on a straight line between centers, which is intuitively obvious.
Maybe it's obvious, maybe it's not. But as you say, his proof does demonstrate that fact.
This is so easy I did this in my head in 30 seconds, 5-1=4, 4-1=3 -> special right triangle 3,4,5 -> 5-2-1=2 -> answer is 2. I'm not a genius but sometimes problems on this channel are easy, and sometimes are really hard. This was the easy one with a misleading "Oxford" caption.
Same here. Why such a long and unnecessary complicated story….
I used this:
-The two closest points always lie on a line that intersects both radii.
-If the two closest points lie on both the line and the circles, then the points must be solutions of both.
-Find the equation of the line using the two radii. y = (3/4)x + 1/4
-Substitute y in each circle’s equation for the line’s equation and solve for x and y. When solving for either variable, choose the point of the two solutions closest to the goal, the two points that would be closest to each other on the circles.
For example, the first circle, (x-1)^2 + ((3/4)x + 1/4 - 1)^2 = 1 yields x = 0.2 and 1.8. I would choose 1.8 since I know the centers of the two circles, (1, 1,) and (5, 4), and would choose the most sensible points to aim for the shortest distance.
-The two points would be (1.8, 1.6) and (3.4, 2.8), having a distance of 2. The key to this question was remembering the property of the closest distance between two circles being on the same line as a line through both diameters. :)
I like how you can use pythagerous for this.
So process is.
1) Draw two circles on a graph at the stated centres and radii.
(x-h)sq +(y-k)sq=rsq
centre=(h,k)
radius=(r)
Circle 1 is origin 5,4 radius 2
Circle 2 is origin 1,1 radius 1
2)Draw a straight line point to point from the centre one circle to the other. Because this is the shortest point between PQ where we know the fixed lengths of both R.
3)Find the length of the line by using pythagerous theorum.
4)Subtract the radii of each circle from the line length.
5)This is the shortest point between the circumfrence of both circles.
6) Lables
H= Hypotenuse, X= horizontal, Y=vetical, rSubscript1= first circle (larger one), rSubscript2=second circle (smaller circle)
P= co-ordinate large circle
Q= co-ordinate small circle
A= Centre of Large Circle
B= Centre of small Circle
7) Vectors which I am not going to simplify as it doesnt lend itself to long writing.
Both are going to use point B as the vector origin.
Vector BA=(4,3) [(BQ)/(BA)=1/5]
Q= B+[1/5 of the vector BA]
Q= 1,1+[1/5 of (4,3)]
Q= 1.8,1.6
Vector BA=(4,3) [BP/BA)=3/5]
Note H minus Rsubscript 1 makes point P the location on the hypotenues length from vector start point B
P= B+[3/5 of vector BA]
P=1,1 +[3/5 of 4,3]
P=3.4, 2,8
Comment added for my own working through the process as a learning tool.
The coordinates of the points are so much easier to find. You can use basic trigonometry.
1. Find the angle of the line ab at point b. You can do this using the right angle triangle shown, with side lengths 4 and 5.
2. Use the angle of the line at point b and the distance between points b and p to create triangle with side lengths that are equal to the coordinates of point p.
3. Repeat for the other point
I didn't remember circle equation based on a grid, but once you gave it, it was very easy to figure out.
The shortest distance lye on line joining centers.Line joining centers is of length √4^2 + 3^2 =5. Shortest distance is 5 - (sum of radii of circles) = 5 -3=2.By forming and solving quadratic equations we get coordinates of P and Q are:(17/5 , 14/5) and (9/5, 8/5,)
gonna start by saying thanks for this problem!
for the coordinates of the points q and p:
first i used y=mx+b to get the equation of that intersecting line (slope can be found using the two center points of the circle)
then set the line equation equal to the equation of the smaller circle to figure out the values of the 2 intersecting x points (and selected the greater of the 2)
then back tracked to find the two intersecting y points of that x value along the equation of the same circle used
from there the same logic could be applied to find out the values for coordinate p
although cumbersome, it was enjoyable to relearn a bunch of old math concepts.
Thanks! 😁
The shortest distance between circles is on the line through both centers which crosses the circles at a right angle. It is the distance between the centers minus the radii of both circles.
From the formulas we know the origins (5,4) and (1,1) and the radii √4 (2) and √1 (1).
The offset between the origins is (4,3) like the sides of the known 3-4-5 triangle so the distance must be the hypotenusa 5. Subtracting the radii 2 and 1 gives 2 as the distance between the circles.
this is the beauty of math, such that you can even do the simplest problems in different ways, the hard and simple problems in simple and hard ways🤷♂
Extremely easy. find distance of centres using distance formula, then subtract sum of radii from that distance
c1=(5,4), r1=2 units; c2=(1,1), r2=1 unit
d=root(((5-1)^2)+((4-1)^2))
=5 units
r1+r2=3 units
shortest distance=5-3=2 units.
The distance between the centers = sqrt((5-1)^2 + (4-1)^2) = 5. Since the radius of the
first circle is sqrt(4) = 2 and the radius of the second circle is sqrt(1) = 1, the circle do not intersect. Hence, the closet two points lie on the line segment joining the two
centers where that line segment intersects the two circles. That distance is, clearly,
5 - (2 + 1) = 2.
Yes, I got same:
circle formula, so:
Circle A is of radius 2 (square root of 4) and centered at (5,4).
Circle B is of radius 1 (square root of 1) and centered at (1,1).
Shortest distance D between the circles is
distance from P to Q,
which is distance between the centers of the two circles,
less the radius of each
D=((5-1)^2+(4-1)^2)^.5-2-1
D=(25)^.5-2-1
D=5-2-1
D=2
Draw line L through the radius of both circles,
it also passes through points P and Q.
L: y=Sx+C
4=5S+C
1=S+C
3=4S
S=3/4
Slope of line is S=3/4
right triangle, 3*3+4*4=5*5, so sides 3, 4, 5
So, for distance 5 along line PQ, x goes up by 4, y goes up by 3,
so, for distance 1 along line PQ, x goes up by 4/5, y goes up by 3/5,
Q is distance radius of B (1) away from center of B (1,1)
P is distance radius of B + D = 3 away from center of B (1,1)
Q: (1+4/5,1+3/5)
P: (1+3*4/5,1+3*3/5)
Q: (1.8,1.6)
P: (3.4,2.8)
For the line between the circles. Connect the circle centers with a line. Determine the length of that line. Subtract from that line the radius of both circles. The result is the length of the shortest line connecting the circles. In this case the line is the hypotenuse of a 3x4x5 triangle. 5 minus 1 minus 2 is 2. The line connecting the circles is length 2.
Centres (5,4) & (1,1). Distance between is 5. (Trivial 345 triangle). Rad 1, rad 2. Shortest distance between circles is 5-2-1=2. Done. No Pythag beyond the 345 triangle. No quadratics.
Just basic geometry. 2:52
Haven't watched the video yet, but my first inclination is actually do this in reverse order:
The shortest distance between two (non-overlapping) circles will always be along the line connecting their two center points, so find the equation for that line, then use the intersection between that line and each circle (two equations in two variables) to find the coordinates for each of P and Q, then use the distance formula to find out the distance between them.
Though I suppose if you didn't need to know P and Q, you could find the distance by just calculating the distance between the centers, and then subtracting the radii of each circle... (might want to do that first anyway just to know that they're actually not overlapping (if the result is negative, they're overlapping, and the answer to the first part is 0))
Circle A has a radius of 2 and circle B has a radius of 1. Circle A’s center is at (5,4) and circle B’s is at (1,1). The distance between these two points is sqrt((4-1)^2 + (5-1)^2) = 5 units. Subtract both radii from this value and you get 2.
I just found the slope between the two centers of the circle, (1,1) and (5,4), using slope intercept form to get 4-1=m(5-1), 3=4m the slope is 3/4, plotting 3/4x we are not at the center, since plugging in 1 would get us 3/4, i add 1/4 to the line, ending up with 3/4x+1/4. Using this we can figure out the edges of the circle by substituting the y in each circle with the equation of line. Set each equation equal to 0, after solving for each solution, you get (0.2,0.4) and (1.8,1.6) for the smaller circle and (3.4,2.8) and (6.6,5.2). We can get rid of each extremes, as (1.8,1.6) and (3.4,2.8) are the closer points, the line connecting them is diagonal, which you could say is a hypotenuse of a triangle, subtracting the coordinates from each, you get a height of 1.6 and a length of 1.2, pythagorean theorem says a^2+b^2=c^2 or the square root of a^2+b^2=c, plugging in 1.6 and 1.2, you get 2.56+1.44, and that equals 4, and then you take the square root, which equals 2. The shortest distance between the two circles is 2.
I completely overcomplicated this by missing the fact that the vector between their centers is composed of the legs of a 3-4-5 triangle. Ended up finding P and Q the hard way first with the angle of this vector and the x-axis (transformed from polar coordinates) then the distance between P and Q. Got the correct answer at least!
@2:05 - it takes 3 seconds to realize that we have right triangle 3-4-5, so the distance is equal to 2.
5 - r1 - r2 = 2
r1=1, r2=2
very conceptual, congratulations!
It’s the distance between the centers, which is 5 ( 3 and 4 are the other sides of a rectangle triangles), minus the 2 radius, so 5-(2+1)=2
This question was really easy. We just need to use distance formula and parametric coordinates of a circle
If we can assume the shortest path is a straight line (which is a well known fact in flat space), it's easy. Draw a straight line from A to Q and call the intersection with the first circle P'. Note that AP and AP' are radii so they have the same length. If also QP and QP' have the same length, the triangle APQ is degenerate and P=P'. Otherwise, QP must be longer than QP' and we can shorten it by replacing P with P'. So If we assume QP is the shortest path, P must already be in the line AQ. Likewise, Q must be on the line BP. So AP, PQ, QB form a straight line and thus P,Q are on AB. The rest is simple calculation, √((4-1)² + (5-1)²) - √4 - √1 = 2.
this ws the first problem on your channel that I was intuitively able to solve in my head lol
The shortest distance will obviously be a straight line from a to b, i did a²+b²=c²
and got PQ=2
For the coordinates
i used trigonometry
With θ and β being angels of the same triangle in Q1
So it will be
Q ⇒ (1+cosθ,1+sinθ) ⇒(1+⅘,1+⅗)⇒(1.8 , 1.6)
P⇒(5-2sinβ,4-2cosβ) ⇒(5-2.⅘ ,4-2.⅗) ⇒(3.4 , 2.8)
Using similarity of the triangles
Also you could just use the triangle inequality to say |AB|
distance between two centers - sum of each circle's radius = 5 - (2 + 1) = 2
I'm definitely not the sharpest tool in the shed. But I still think that the approach you took to solve the first part of the problem is unnecessarily complicated.
Yes, there is no need to go into the optimization digression. APQB is a quadrilateral, so AB is smaller than the sum of the other three sides. But still an awesome video.
Well he was trying to prove the result, not just get the result.
I used the slope of the line through their midpoints a tiny touch of trig to add/subtract to/from the coords of the midpoints. It was made trivial by the fact that it was a 3,4,5 triangle. Your way was a little cumbersome.
I got the answer (2) in my head in about four seconds. The centerpoints of the circles make a 3-4-5 right triangle, the radii are 1 and 2, so 5-1-2 = 2.
for the second question, I made a function for the hypotenuse (the line of the shortest distance between the two centers of the circles) and just used abit of trig to get the x values of q and p
If the circles are not intersecting, then those two points lies on the line that goes through their centres. We can create an equation for such a line and that gives us two points for each circle. We then pick the points that are nearest to each other. These two points also answer the first question, because their distance is also the shortest distance between those two circles, so you can solve second question and get answer to the first, or you can solve the first by calculation of distance of their centers and then subtracting the radius.
I solved this in like 2 minutes (toke the distance from center of circle A to center of circle B (hypotenuse of a right angle triangle) and substracted radius a ad radius B.... (didn't bother with part 2, no fun)
I then watched the video and wondered why on earth the solution was strating out so complicated... honestly dropped of before the end
The best way to teach math is to keep it simple
The shortest distance D is the length of the part P of the line L connecting the centers where P is measured from the intersections btw. L with A and B respectively. The length of L is the distance of the centers of CA = (5,4) and CB = (1,1) which is sqrt(4^2+3^2) = 5. The two radius RA=sqrt(4)=2 and RB=sqrt(1)=1 at both ends of L are not part of P so they must be subtracted from lengt od L. So length of P = D = 5-2-1 = 2. Takes no 10 seconds to do this in head.
It’s on a graph just create a right angle triangle and you see it has sides of 4 and 3 between the 2 centres with the hypothenuse being the min distance of 5.
Part 1:
Those equations make it very straightforward. The first circle has center (5,4) and the second (1,1). The distance between those two points is 5. The radii of the two circles are 2 and 1 respectively, so that means the shortest distance between two points on the two circles is 5-2-1=2.
Part 2:
To go from the center of A to B, you have to follow a line that goes 4 left and 3 down. If you follow 2/5th of that line, you land on point P, which is (5-4*2/5,4-3*2/5) = (3 + 2/5, 2 + 4/5). If you follow 4/5th of that same line, then you land on point Q, which is (5-4*4/5,4-3*4/5) = (1 + 4/5, 1 + 3/5).
The shortest distance between centers is the hypothenus, move B(1,1) to (5,4) and the shortest distance 0f the perimeters i 0
It took 4 minutes to explain that the distance between two points is 5, and the radii around those points are total length 3. The second solution was interesting; I used trig ratios, but I like how it was done here.
What an ungainly solution to a fairly simple problem?
Part A: Line between centres of circles must give the shortest distance because radii cut at 90 degrees. Shortest is simply pythagoras - total radii
Part B is just similar triangles
For the distance to the circles, just make a line using the centers and then subtract the radii.
I got 2
Got it by the 17-sec mark of the video! Hope I’m right!!
For part 2 of the exercise, you can take the equation for line AB, and simply calculate the coordinates where the line intersects each circle.
Took about 2 seconds to figure it out. I should go to Oxford. Great little brain teaser.
15 seconds to solve.
1) Center of the first circle (x; y = 5; 4)
Center of the second circle (x; y = 1; 1)
2) Distance between centers:
√((5-1)^2 + (4-1)^2) =
√(4^2 + 3^2) =
√(16 + 9) = √25 = 5
3) Subtract the radii of the circles from the resulting distance, and get the shortest distance: 5 - 2 - 1 = 2
Answer: shortest distance is 2
How I solved part 1: Shortest distance between 2 points is a straight line, so points P and Q will be along Segment AB. A line down from point A and a line right from point B will intersect at 5,1 and form a right angle. Segment AB then forms the hypotenuse and makes this a right triangle whose sides are 3 and 4...this is a 3-4-5 right triangle. Segment AB is 5 units long. Circle A's radius is sqrt(4), so it's 2. Circle B's radius is sqrt(1), so it's 1. 5 - 2 - 1 = 2, so Segment PQ will be 2 units.
As soon as I saw the problem I instantly though, "lmao we have the distance formula for a reason, just subtract the radius" like idk why this is such a "hard" problem.
You can figure the centre and radius out by the equation itself and then calculate the distance from the centres then subtract the sum of both radii
That'll give you the shortest distance
If the circles are plotted corrrectly, the distance from A to B is the length of the hypotenuse of a 3,4,5 triangle, so that distance is 5. Now we need only subtract the length of the two radii to get the distance between the circles: 5 - 2 - 1 = 2.
Nice problem! Managed to solve both parts with a pencil and paper, no calculator.
The distance between 2 circles (or any shapes) depends on where they are placed relative to each other! If you place them 1cm apart, then that's the distance between them. If you place them touching, then the distance is zero. So zero is the shortest distance between 2 circles...
If you know how circle equations work and know the 3-4-5 triangle, you can literally solve this in seconds. I did!
If only this was the actual question I got... Immediately clear what to do and not too much work to get there.
A circle radius 2 at 5,4, and radius 1 at 1,1. Translate the 1,1 to 0, and the centre at that first circle is at 4,3. So, a pythagorean triple. Duh. Distance between the centres is 5, subtract the two radii (1 and 2), and we are left with a distance of 2.
Let's watch the video and see how that goes.
--edit--
oh, there's a second part. Looking at the line through the centers, we have that pythgorean triad triangle again inside each circle, the hypotenuse being the radius. So the x coordinates will be 1+4/5, 1+3/5 = (1.8, 1.6), and the coordiates of the second will be 5-2*4/5, 4-2*3/5, or (3.4, 3.8).
Lots of people more interested in saying they know how to do this, and uh way easier akshully, than people recognizing the point he's trying to make (how to think through a math problem). Its a lot less about bragging on right answers and more about teaching people how to think. This was a GREAT video for explaining to my 6th grader how the lesson is knowing how to combine simple math facts to think through a math problem.
I thought you made that explanation more difficult than it needed to be.
i never would have known until about 5 months ago when I started looking into ray marching
Now I felt smart when I found the solution obvious. 😊
The use of the expression "Daunting Problem" had me spending far too long thinking I had misunderstood the question, but no! it really was a straight line between the centres. Once I realised that I was thinking too hard it was easy and so was the rest of it. Is this an indictment of the university system or does "Oxford admissions" refer to the McDonalds in Oxford?
2:07 The equations would give you the respective radius, but what gives you the coordinates of their centres? Can they not be tangent?
I only watched up to 2m20s. The 2 centres are corners of a triangle and we know the vertical and horizontal lengths so we know the diagonal length ie distance between centres. Then subtract the radius of each circle
The whole line drawing and point moving along the circle was harder to do than the whole math on this one. ;) Got the first part on looking at the question and the second one easily after. That said, i got a masters degree in maths and so this SHOULD be easy. 😅
How to construct a PhD thesis out of a simple question, the most Oxford thing ever
He gives an interesting proof, did you see that?
Oh, this is hard, I thought. Then I thought about it for a minute and then I solved it in my head in 10 seconds.
Are they on the same plane? Fold the plane until the circles lay on each other.
They are on the same plane as nothing references the Z axis, only X and Y axis’ are referenced.
I don't understand how it isn't obvious that the shortest distance lies on a line cutting through the middle of both circles.
If the circles were on top of each other, vertical rather than on a diagnal, then it would be SUPER obvious because any other place on the circle would be further away, because the points on the line would be the highest or lowest point on each circle.
Sadly, it gets more complocated trying to explain it than just looking at it.
And calculating PQ is pretty easy after that.
This reminds me of a problem in highschool Algebra where the teacher drew an isosceles(?) triangle and labeled the long side as 7 and the two short sides as 3.
She asked us to give her the internal angles. One look, and I determined it was impossible because 3 + 3 = 6, which is less than 7, so this is an imoossible triangle.
I got it wrong for not showing my work using trigonometry. Sometimes, logic is best. 😅
For me, part 1 use 5-2-1=2 , and part 2 use point of intersection formula x-coor of P=(3*5+2*1)/(3+2) and so on...
It's 2 am and I saw this ... Solved it in under a min (yea)
Every circle equation is in the format (x-h)^2 + (y-k)^2 =r^2
Where the centre of the circle is (h,k) and radius is r
After finding centres of both circles use distance formula to find distance √{(x-a)^2 + (y-b)^2} where (a,b) is centre of other circle.. now subtract sum of radiii of both circles from the distance calculated above and tada...
All this is taught to 15 year olds in India who are preparing for IIT (a super difficult exam and I am preparing for it)
And you didn't get part 2.... and hence, failed.
@@turbodog99 I didn't try.
Man, solved this one in 15 seconds in my head :)
They are circles, so the distance betwen them is the distance between the centers minis the sum of the radii.
I have to admit I easily figured it out once the circles were shown on the graph, but when he referred to equations for circles, I had no idea what he was talking about until he showed it. In hindsight, I probably did learn it in school, but it did not stick with me at all.
Pitagoras 3,4 =>5 5-2-1= 2