Find the equation of a circle which touches both the axes and the line $3x-4y+8=0$ and lies in the third quadrant.

Question

[Hint : Let $a$ be the radius of the circle ,then $(-a,-a)$ will be centre and perpendicular distance from the centre to the given line gives the radius of the circle.]

$\begin{array}{1 1}x^2+y^2+4x+4y+4=0\\x^2+y^2-4x-4y+4=0\\x^2+y^2-4x-4y-4=0\\x^2+y^2+4x+4y-4=0\end{array}$