Question

已知圆与x轴相切,圆心在直线x-2y=0上,且这个圆经过a(4,1)点,求该圆的方程

Answer 1

圆心在直线x-2y=0上,设圆心坐标是(2m,m),又与X轴相切,则半径r=|m|.

那么圆方程是:(x-2m)^2+(y-m)^2=m^2

(4,1)代入得:(4-2m)^2+(1-m)^2=m^2

16-16m+4m^2+1-2m+m^2=m^2

4m^2-18m+17=0

m=(18(+/-)2根号13)/8=(9(+/-)根号13)/4.

代入上面方程即可.有二个.