与圆x^2+y^2+6x-4y+4=0切于点A（-6/5,-2/5）,且和直线x=2相切的圆的方程为

Question

与圆x^2+y^2+6x-4y+4=0切于点A（-6/5,-2/5）,且和直线x=2相切的圆的方程为

Answer 1

圆x^2+y^2+6x-4y+4=0(x+3)^2+(y-2)^2=9圆心是(-3,2),半径是r=3设所求圆为(x-a)^2+(y-b)^2=r^2(r＞0)则(a+3)^2+(b-2)^2=(r+3)^2.(1)(-6/5-a)^2+(-2/5-b)^2=r^2.(2)2-a=r.(3)联立(1)(2)(3)式得a=0,b=-2,r=2所以所求圆...