Find the equation of the circle whose center is $\left(6, 8\right)$ and which passes through the center of the circle $x^2 + y^2 + 12x + 8y = 48$
Equation of given circle is
$x^2 + y^2 + 12x + 8y = 48$ $\;$ i.e. $\;$ $x^2 + y^2 + 12x + 8y - 48 = 0$
Comparing with the standard equation of the circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives
$g = 6$, $\;$ $f = 4$, $\;$ $c = -48$
Center of the given circle $= \left(-g, -f\right) = \left(-6, -4\right)$
The required circle has center at $\left(6, 8\right)$ and passes through the point $\left(-6, -4\right)$
$\therefore \;$ Radius of the required circle $\;$ $= \sqrt{\left(6 + 6\right)^2 + \left(8 + 4\right)^2} = \sqrt{288}$
$\therefore \;$ Equation of the required circle is
$\left(x - 6\right)^2 + \left(y - 8\right)^2 = \left(\sqrt{288}\right)^2$
i.e. $\;$ $x^2 -12x + 36 + y^2 - 16y + 64 = 288$
i.e. $\;$ $x^2 + y^2 -12x - 16y - 188 = 0$