求圆心在直线3x+y=0上,过原点且被y轴截得的弦长为6的圆的方程?回答呢?

Question

求圆心在直线3x+y=0上,过原点且被y轴截得的弦长为6的圆的方程?
回答呢?

Answer 1

可设圆心O为(m,-3m); 方程为(x-m)^2+(y+3m)^2=R^2
圆过原点,故R^2=m^2+(-3m)^2=10m^2;
即圆为(x-m)^2+(y+3m)^2=10m^2;
Y轴截得的弦长为6,故圆方程中,令x=0 易得y1=-6m, y2=0,舍去
故|-6m|=6,即m=土1.因此,圆方程为(x-1)^2+
(y+3)^2=10,或(x+1)^2+(y-3)^2=10.