Question

已知圆C的圆心是直线x-y+1=0与x轴的交点,且圆C与直线x+y+3=0相交于A,B两点,|AB|=2根号2,求圆的方程|

Answer 1

圆C的圆心在（-1 ，0），设圆的方程为
(x+1)^2 + y^2 = r^2
即 x^2 + 2x + y^2 + 1 - r^2 = 0
设它与直线x+y+3=0的交点为(x1,y1)，(x2,y2)
将x+y+3 =0 代入圆方程有
2x^2+8x+10-r^2 =0
Δ=64 - 8(10-r^2)
则
y1 = -(x1+3)
y2 = -(x2+3)
(y2-y1)^2 + (x2-x1)^2
= (2√2)^2 = 8
(y2-y1)^2 + (x2-x1)^2
= 2(x2-x1)^2
= 2(2√Δ/4)^2
= Δ/2
= 32 - 4(10-r^2)
= 4r^2 - 8
所以
4r^2 - 8 = 8
r^2 = 4
所以圆的方程为
(x+1)^2 + y^2 = 4

Answer 2

(x+1)二次方+y二次方=4