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

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