Equation of the circle tangent to xy40 and 7xy40 whose cente

Equation of the circle tangent to x+y+4=0 and 7x-y+4=0 whose center on the line 4x+3y-2=0.

In finding the equation of circle, we need radius and center coordinates of the circle.

First given the tangents x+y+4=0 and 7x-y+4=0

so intersection point of those two is (-1,-3) .

now line CP is the bisector of angle APB so applying angle(APC) = angle(CPB)

we will get |(-1-m)/(1-m)| = |(m-7)/(1+7m)|

so solving this we will get m={1/3 , -3}

so equation of line CP can be 3h+k+6 = 0; h-3k-8=0

so finding the intersection point of each of this line with the given line 4h+3y-2 = 0 gives the coordinates of the center.

By solving fisrt possible CP equation and given equation we get center as (-4,6)

and with second we get (2,-2)

so radius of the circle is the distance from center to any of the tangent given

that is r= 2sqrt(2) when (2,-2) is the center and 3sqrt(2) when (-4,6) is the center.

so two possible equations of the circle are (x-2)^2 + (y+2)^2 = 8

and (x+4)^2 + (y-6)^2 = 18.