By symmetry, another line can be drawn parallel to CD on the left at a distance 2. This makes a rectangle of sides (16, 2) with a diagonal √260 = 2R R = √65
By intersecting chords theorem , the upper segment of vertical chord is 6•8/12=4 Drop a perpendicular line from centre of circle to vertical chord, then you can see a right triangle of side length of 1 and 8 with radius of circle being the hypotenuse . r ^2 = 8•8 + 1•1 r = sqroot of 65
Another method. First, the distance from P to C =4 (intersecting chord theorem, since 8*6= 12*4) Could you draw a line ( the diameter) through the circle's center to bisect the chord? So, the distance from the center of the chord is 7. The Pythagorean theorem can be used to calculate the distance from the chord's center to the circle's edge. Since the hypotenuse from the center of the chord to C i =sqrt (1^2 + 4^2)= sqrt 17 or 4.123 , then the distance from the center of the chord to the edge of the circle is 4 since it has the same hypotenuse 4.123 and the same base 1 Draw a line from the center of the circle ) to A. This is R, and from A to the center of the chord is 7. From the center of the chord to the center of the circle, the is distance is R-4, Using Pythagorean with the sides R, R-4, and 7 will give R R^2 = (R-4)^2 + 7^2 R^2 = R^2 + 16 - 8R + 49 0 = 16 -8R + 49 8R = 16 + 49 8R = 65 R = sqrt 65 =8.0622
I've actually encountered a similar large measurement problem in real life, so this caught my interest. Unfortunately, the intersecting chords must be exactly perpendicular to each other, which is not easy to guarantee in reality. So disappointingly this doesn't work. (The way it was measured in practice, was finding center by an iterative method of crossing lines, and then measure from center to edge.) Taking a wild chance, I examined if the second method would work with non-perpendicular lines. Nope, it won't, pity.
Another possibility is to construct a triangle CAE such that CE passes through the center, and thus ∠CAE = 90° If ∠AEC = θ, then ∠ADC = θ since they share the same chord AC. In ∆CAE, Sinθ= AC/2R = √80/(2R) In ∆DPA, Tanθ = AP/DP = 8/12 = 3/4 Hence, Sinθ = 2/√13 R = √65
Potencia del punto P respecto a la circunferencia: 6×8=12×CP》CP=4 》 AB=8+6=14=7+7 》CD=4+12=16=8+8》Su punto medio dista 12-8=4 unidades de la cuerda AB 》 Potencia del punto medio de AB respecto a la circunferencia: 7×7=(r-4)(r+4)》 r=sqrt65 Gracias y un saludo.
I solved the problem by drawing perpendicular lines through the diameter and then used the intersecting cords theorem to find r. My first equation came from drawing a diameter perpendicular to line segment CD, which gave me an equation of (r+1)(r-1) = (8)(8); r^2 -1 = 64; r^2 = 65; r = sq rt(65). My second equation came from drawing a diameter perpendicular to line segment AB, which game me an equation of (r+4)(r-4) = (7)(7), r^2 -16 = 49; again, r^2=65; r= sq rt(65). That is how I did it. Thanks for the problem.
∠APD=90° so I think AB as x-axle,CD as y-axle. Given O(x, y), by applying the Pythagorean theorem, r^2=(6+x)^2+y^2=(8-x)^2+y^2=x^2+(12-y)^2 (6+x)^2+y^2=(8-x)^2+y^2 6^2+2*6*x+x^2+y^2=8^2-2*8*x+x^2+y^2 6^2+2*6*x=8^2-2*8*x 36+12x=64-16x (12+16)x=64-36 28x=28 x=1 (6+x)^2+y^2=x^2+(12-y)^2 (6+1)^2+y^2=1^2+(12-y)^2 7^2+y^2=1^2+(12-y)^2=1^2+12^2^2*12*y+y^2 7^2=1^2+(12-y)^2=1^2+12^2-2*12*y 49=1+144-24y 24y=1+144-49=96 y=4 r^2=(6+x)^2+y^2=(6+1)^2+4^2=7^2+4^2=49+16=65 r=√65
Another solution, but it did work. Join AD & DB. Tan ADP = 8/12. Angle ADP = (tan -1) 8/12 = 33.690 degrees. Tan PDB = 6/12. Angle PDB = (tan-1) 6/12 = 26.565 degrees. Thus angle ADB = 33.690 + 26.565 = 60.255 degrees. Now the angle at the centre is twice the angle at circumference. Joining AO & OB, angle AOB = 2 x 60.255 = 120.510. Then angles OAB & OBA = (180 - 120.510) / 2. Angles OAB & OBA are 29.745 each. Applying the sine rule to triangle AOB. 14 / sin 120.510 = r / sin 29.745. r = 14 sin 29.745 / sin 120.510. r = 8.062.
I unintentionally made it difficult for myself (as usual), but I will share my odd method anyway as it may possibly have some merit. I figured out the unlabelled line was 4 due do multiplying the line segments, but then I thought, "What if I move the vertical line across by 1 to form a diameter with a length of 7 on each side?" That would make the multiplication 49 (7*7). This would make CD slightly longer so I then did (12+x)(4+x)=49. After converting to x^2+16x-1=0 and employing the quadratic formula, I eventually came up with a solution for x which led to the radius. I used a calculator too.
If the coordinate of the point p be (0,0) Then the coordinate of the points A, B, D will be (6,0) (-8,0) (0,-12) Let (p, q ) be the center and r the radius of the circle So (-6-p) ^ 2 + (0-q)^2 =r ^2 (-8-p)^ 2 + (0 - q) ^2 = r ^2 ( 0 - p)^ 2 + (-12 - q )^2 =r ^ 2 By solving this 3 equations The center of the circle is (-1,-4) And the radius is sqrt(65)
Intersecting chords theorem: a. 12 = 6 . 8 a = 4 cm Again same theorem: (R+1)(R-1)= (½(12+4))² R² - 1 = 8² R² = 64 + 1 = 65 R = √65 = 8,06 cm ( Solved √ ) Again the same in the other direction: (R+4)(R-4)= (½(8+6))² R² - 16 = 7² R² = 49 + 16 = 65 R = √65 = 8,06 cm ( Solved √ )
Solution: According to the principle of chords: CP*PD = AP*PB ⟹ CP = AP*PB/PD = 8*6/12 = 4 r = OD = √[(CD/2)²+(AB/2-PB)²] = √[8²+(7-6)²] = √65 ≈ 8.0623
if we draw a parallel line for CD which pass the point O, we can get upper side of AB for (r-4), and lower side for (r+4). next step is similar to first method.
Let x be the upper width of the vertical chord, so half of its length is (12+x)/2=6+x/2, and half of the length of the horizontal chord is 7, now let r be the radius of the circle, so r^2=1+(6+x/2)^2=49+(6+x/2-x)^2=49+(6-x/2)^2, thus 48=(6+x/2)^2-(6-x/2)^2=12x, so x=4, therefore r=sqrt(1+8^2)=sqrt(65).😅
Thanks a lot master.....
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@@PreMath I am learn more about math at your channel ....
By symmetry, another line can be drawn parallel to CD on the left at a distance 2. This makes a rectangle of sides (16, 2) with a diagonal √260 = 2R
R = √65
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By intersecting chords theorem , the upper segment of vertical chord is 6•8/12=4
Drop a perpendicular line from centre of circle to vertical chord, then you can see a right triangle of side length of 1 and 8 with radius of circle being the hypotenuse .
r ^2 = 8•8 + 1•1
r = sqroot of 65
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Another method.
First, the distance from P to C =4 (intersecting chord theorem, since 8*6= 12*4)
Could you draw a line ( the diameter) through the circle's center to bisect the chord? So, the distance from the center of the chord is 7. The Pythagorean theorem can be used to calculate the distance from the chord's center to the circle's edge. Since the hypotenuse from the center of the chord to C i =sqrt (1^2 + 4^2)= sqrt 17 or 4.123
, then the distance from the center of the chord to the edge of the circle is 4 since it has the same hypotenuse 4.123 and the same base 1
Draw a line from the center of the circle ) to A. This is R, and from A to the center of the chord is 7. From
the center of the chord to the center of the circle, the is distance is R-4,
Using Pythagorean with the sides R, R-4, and 7 will give R
R^2 = (R-4)^2 + 7^2
R^2 = R^2 + 16 - 8R + 49
0 = 16 -8R + 49
8R = 16 + 49
8R = 65
R = sqrt 65
=8.0622
Thanks, dear
I've actually encountered a similar large measurement problem in real life, so this caught my interest. Unfortunately, the intersecting chords must be exactly perpendicular to each other, which is not easy to guarantee in reality. So disappointingly this doesn't work. (The way it was measured in practice, was finding center by an iterative method of crossing lines, and then measure from center to edge.) Taking a wild chance, I examined if the second method would work with non-perpendicular lines. Nope, it won't, pity.
Another possibility is to construct a triangle CAE such that CE passes through the center, and thus
∠CAE = 90°
If ∠AEC = θ, then ∠ADC = θ since they share the same chord AC.
In ∆CAE, Sinθ= AC/2R = √80/(2R)
In ∆DPA, Tanθ = AP/DP = 8/12 = 3/4
Hence, Sinθ = 2/√13
R = √65
Excellent!
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8/12= 2/3 (tan < DPA), but we don't need this value since AD= 4 rad (13) by PT, then use sine ratio as you did above. Good job!
@@octaviohernandez18Right :)
Pēc Pitagora teorēmas BD=6*sqr5, AD=4*sqr13 , ABD laukums:S=14*12/2=84 ,trīsstūrī ABD R=a*b*c/4S=6*sqr5*4*sqr13*14/84=sqr65.
Potencia del punto P respecto a la circunferencia: 6×8=12×CP》CP=4 》 AB=8+6=14=7+7 》CD=4+12=16=8+8》Su punto medio dista 12-8=4 unidades de la cuerda AB 》 Potencia del punto medio de AB respecto a la circunferencia: 7×7=(r-4)(r+4)》 r=sqrt65
Gracias y un saludo.
Thanks, dear
I solved the problem by drawing perpendicular lines through the diameter and then used the intersecting cords theorem to find r. My first equation came from drawing a diameter perpendicular to line segment CD, which gave me an equation of (r+1)(r-1) = (8)(8); r^2 -1 = 64; r^2 = 65; r = sq rt(65). My second equation came from drawing a diameter perpendicular to line segment AB, which game me an equation of (r+4)(r-4) = (7)(7), r^2 -16 = 49; again, r^2=65; r= sq rt(65). That is how I did it. Thanks for the problem.
Super!
∠APD=90° so I think AB as x-axle,CD as y-axle.
Given O(x, y), by applying the Pythagorean theorem,
r^2=(6+x)^2+y^2=(8-x)^2+y^2=x^2+(12-y)^2
(6+x)^2+y^2=(8-x)^2+y^2
6^2+2*6*x+x^2+y^2=8^2-2*8*x+x^2+y^2
6^2+2*6*x=8^2-2*8*x
36+12x=64-16x
(12+16)x=64-36
28x=28
x=1
(6+x)^2+y^2=x^2+(12-y)^2
(6+1)^2+y^2=1^2+(12-y)^2
7^2+y^2=1^2+(12-y)^2=1^2+12^2^2*12*y+y^2
7^2=1^2+(12-y)^2=1^2+12^2-2*12*y
49=1+144-24y
24y=1+144-49=96
y=4
r^2=(6+x)^2+y^2=(6+1)^2+4^2=7^2+4^2=49+16=65
r=√65
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Thank you!
You're welcome!
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Nice formula you have developed sir, does it work on all circles in finding radius and diameter using the chord theorem?
Yes, as long as they're perpendicular. Cheers
Beautiful steps for solving the problem
Can you proof this formula
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Sure, pretty soon!
Another solution, but it did work.
Join AD & DB.
Tan ADP = 8/12.
Angle ADP = (tan -1) 8/12 = 33.690 degrees.
Tan PDB = 6/12.
Angle PDB = (tan-1) 6/12 = 26.565 degrees.
Thus angle ADB = 33.690 + 26.565 = 60.255 degrees.
Now the angle at the centre is twice the angle at circumference.
Joining AO & OB, angle AOB = 2 x 60.255 = 120.510.
Then angles OAB & OBA = (180 - 120.510) / 2.
Angles OAB & OBA are 29.745 each.
Applying the sine rule to triangle AOB.
14 / sin 120.510 = r / sin 29.745.
r = 14 sin 29.745 / sin 120.510.
r = 8.062.
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Thanks for video.Good luck sir!!!!!!!!!
Excellent!
You are very welcome!
Thank you! Cheers! 😀
I unintentionally made it difficult for myself (as usual), but I will share my odd method anyway as it may possibly have some merit.
I figured out the unlabelled line was 4 due do multiplying the line segments, but then I thought, "What if I move the vertical line across by 1 to form a diameter with a length of 7 on each side?" That would make the multiplication 49 (7*7). This would make CD slightly longer so I then did (12+x)(4+x)=49. After converting to x^2+16x-1=0 and employing the quadratic formula, I eventually came up with a solution for x which led to the radius. I used a calculator too.
🙏🏻 good morning sir, your teaching proccess always motivate us thanks
6*8=12*b
b=4
Centre=((-8+6)/2,(-12+4)/2)=(-1,-4)
Radius=sqrt(1^2+8^2)=sqrt(65)
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If the coordinate of the point p be (0,0)
Then the coordinate of the points A, B, D will be (6,0) (-8,0) (0,-12)
Let (p, q ) be the center and r the radius of the circle
So (-6-p) ^ 2 + (0-q)^2 =r ^2
(-8-p)^ 2 + (0 - q) ^2 = r ^2
( 0 - p)^ 2 + (-12 - q )^2 =r ^ 2
By solving this 3 equations
The center of the circle is (-1,-4)
And the radius is sqrt(65)
Next video will be very exciting. 🙂
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Sir where are you from
Intersecting chords theorem:
a. 12 = 6 . 8
a = 4 cm
Again same theorem:
(R+1)(R-1)= (½(12+4))²
R² - 1 = 8²
R² = 64 + 1 = 65
R = √65 = 8,06 cm ( Solved √ )
Again the same in the other direction:
(R+4)(R-4)= (½(8+6))²
R² - 16 = 7²
R² = 49 + 16 = 65
R = √65 = 8,06 cm ( Solved √ )
Solution:
According to the principle of chords: CP*PD = AP*PB ⟹
CP = AP*PB/PD = 8*6/12 = 4
r = OD = √[(CD/2)²+(AB/2-PB)²] = √[8²+(7-6)²] = √65 ≈ 8.0623
CP = 4. CD = 16. Rsq = 1 + 8sq So R = 8.06
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The first but cumbersome method is to determine the radius of the circumcircle of ABD. but the second method ?
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Just similar triangle with angles in the same segment
3rd method: (r-4)*(r+4)=7^2...?
4th method: (r-1)*(r+1)=8^2...
Very good!
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How do you come out this method?
Sir .Give us explanation please
if we draw a parallel line for CD which pass the point O, we can get upper side of AB for (r-4), and lower side for (r+4). next step is similar to first method.
3rd method please
Please proof the intersecting cord theorem
Hello dear, I've already uploaded the video:
th-cam.com/video/lw70CyiEpmM/w-d-xo.html
Cheers
2nd methed is very interesting and easy.
Intersecting chords theorem:
a . b = c. d
a. 12 = 6 . 8
a = 4 cm
Imagine a quarter circle:
Chord² = R² + R² = 2R²
R² = ½ Chord²
Now, imagine the whole circle:
R²= ½ Chord²
Also :
R² = (½ C² + ½ C²)/2
R² = ¼ C² + ¼ C²
R² = ¼ ( C² + C² )
Also:
R² = ¼ (a² + b²) + ¼ (c²+d²)
R² = ¼ ( a² + b² + c² + d²)
R² = ¼ (4² + 12² + 6² + 8²)
R² = 65
R = √65 = 8,06 cm (Solved √ )
Let x be the upper width of the vertical chord, so half of its length is (12+x)/2=6+x/2, and half of the length of the horizontal chord is 7, now let r be the radius of the circle, so r^2=1+(6+x/2)^2=49+(6+x/2-x)^2=49+(6-x/2)^2, thus 48=(6+x/2)^2-(6-x/2)^2=12x, so x=4, therefore r=sqrt(1+8^2)=sqrt(65).😅
Proposition 11 Archimedes' book of lemmas
Nice! r= √(64 + 1)
Excellent!
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arccos7/r+arccossqrt180/2r=arctg6/12...r^2=50
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@@PreMath ho sbagliato ... è arctg12/6....r^2=65
I thought you said that you'd show the proof of r²=(a²+b²+c²+d²)/4 in the following video?
Avec le rayon OD, r^2 = 8^2 + 1^2 😅
Please the proof that r² = (a²+b²+c²+d²)/4
❤😊😊
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you don't have to keep repeating the same theorem, we can remember what you said 10 seconds ago
asnwer=12cm isit
Intersecting chords theorem:
a. 12 = 6 . 8
a = 4 cm
Half chord y= (12+4)/2 = 8
Half chord x= (8+6)/2 = 7
Taking the appropriate right triangle:
R² = 7²+4²
also
R² = 8²+1²
R = √65 = 8,06 cm ( Solved √ )
😀👍🍺
Excellent!
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@@PreMath Thanks Professor!❤️