HKDSE 2014 Q42. centre:-1,2 radius:2 pt A or B can be (x,x-k) sub (x,x-k)→ (x+1)²+(x-k-2)²=4 turn it into ax²+bx+c=0 form then answer= -b/2a similar as your process on the other hand, there is a simpler way: as slope of x-y=k is 1 slope of line that touch both centre and mid pt of AB=-1 ( a property of circle, i forgot the abb) that mid pt be x,x-k, thus: (2-x+k)/(-1-x)=-1 2-x+k=1+x x=(1+k)/2
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HKDSE 2013 Q.42
如果識得點與直線距離的公式就比較簡單。
圓心是（-1，1）半徑是 3. 因為要相交， 圓心與直線距離
HKDSE 2014
Q42.
centre:-1,2
radius:2
pt A or B can be (x,x-k)
sub (x,x-k)→ (x+1)²+(x-k-2)²=4
turn it into ax²+bx+c=0 form
then answer= -b/2a similar as your process
on the other hand, there is a simpler way:
as slope of x-y=k is 1
slope of line that touch both centre and mid pt of AB=-1 ( a property of circle, i forgot the abb)
that mid pt be x,x-k,
thus:
(2-x+k)/(-1-x)=-1
2-x+k=1+x
x=(1+k)/2
It is a good method. But I would suggest not using variable x to represent the coordinates of the mid-point. It is confusing.
Thanks!