问题描述: 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| 1个回答 分类:综合 2014-11-13 问题解答: 我来补答 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 =0Δ=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 = 8r^2 = 4所以圆的方程为(x+1)^2 + y^2 = 4 展开全文阅读