设圆C上的点A(2,3)关于直线X+2Y=0的对称点仍在圆上,且圆与直线X-Y+1=0相交的弦长2根号2,求圆的方程.

圆C上的点A(2,3)关于直线x + 2y =0的对称点(A;)仍在圆上, 显然x + 2y = 0平分圆,即C在x + 2y = 0上, 设C(-2b, b), 半径r, r² = AC² = (2 + 2b)² + (3 - b)² (i)
圆与直线x - y+1=0相交的弦长2√2, 半弦长p =√2
C与直线x - y+1=0的距离d = |-2b - b + 1|/√2 = |3b - 1|/√2
显然r² = d² + p²
(2 + 2b)² + (3 - b)² = (3b - 1)²/2 + 2
b² + 10b + 21 = 0
b = -3, C(6, -3), r² = 52, 圆的方程: (x - 6)² + (y + 3)² = 52
或
b = -7, C(14, -7), r² = 244, 圆的方程: (x - 14)² + (y + 7)² = 244