Find the equation of the circle concentric with the circle $\;$ $x^2 + y^2 - 6x + 12 y + 15 = 0$ $\;$ and of double its radius.
Given equation of circle: $\;\;\;$ $x^2 + y^2 - 6x + 12 y + 15 = 0$
Comparing with the standard equation: $\;\;\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives
$2g = -6 \implies g = -3$; $\;$ $2 f = 12 \implies f = 6$; $\;$ $c = 15$
$\therefore$ $\;$ Center of given circle $= C = \left(-g, -f\right) = \left(3, -6\right)$
Radius of given circle $= r = \sqrt{g^2 + f^2 - c} = \sqrt{\left(-3\right)^2 + \left(6\right)^2 - 15} = \sqrt{30}$
$\because$ $\;$ The required circle is concentric with the given circle,
center of required circle $= C = \left(3, -6\right)$
Radius of required circle $= R = 2r = 2 \sqrt{30}$
$\therefore$ $\;$ Equation of required circle is:
$\left(x - 3\right)^2 + \left(y + 6\right)^2 = 120$
i.e. $x^2 + y^2 - 6x + 12 y + 9 + 36 - 120 = 0$
i.e. $x^2 + y^2 - 6x + 12y - 75 = 0$