Find the center and radius of the circle whose equation is $\;$ $\left(\ell x + my + 3 \ell\right)^2 + \left(mx - \ell y - 3m\right)^2 = 24 \left(\ell^2 + m^2\right)$
Given equation of circle: $\;$ $\left(\ell x + my + 3 \ell\right)^2 + \left(mx - \ell y - 3m\right)^2 = 24 \left(\ell^2 + m^2\right)$
i.e. $\;$ $\ell^2 x^2 + m^2 y^2 + 9 \ell^2 + 2 \ell m x y + 6 \ell m y + 6 \ell^2 x$
$\hspace{1cm}$ $+ m^2 x^2 + \ell^2 y^2 + 9 m^2 - 2 \ell m x y + 6 \ell m y - 6 m^2 x = 24 \ell^2 + 24 m^2$
i.e. $\;$ $\left(\ell^2 + m^2\right)x^2 + \left(\ell^2 + m^2\right)y^2 + \left(6 \ell^2 - 6 m^2\right)x + 12 \ell m y - \left(24 \ell^2 + 24 m^2\right) = 0$
i.e. $\;$ $x^2 + y^2 + \left[\dfrac{6\left(\ell^2 - m^2\right)}{\ell^2 + m^2}\right]x + \left[\dfrac{12 \ell m}{\ell^2 + m^2}\right]y - 24 = 0$
Comparing with the standard equation of circle $\;$ $x^2 + y^2 + 2 gx + 2fy + c = 0$, $\;$ we have
$g = \dfrac{3 \left(\ell^2 - m^2\right)}{\ell^2 + m^2}$, $\;$ $f = \dfrac{6 \ell m}{\ell^2 + m^2}$, $\;$ $c = -24$
center of the circle $= \left(-g, -f\right) = \left(\dfrac{-3 \left(\ell^2 - m^2\right)}{\ell^2 + m^2}, \dfrac{-6 \ell m}{\ell^2 + m^2}\right)$
radius of the circle $= r = \sqrt{g^2 + f^2 - c}$
i.e. $\;$ $r = \sqrt{\left(\dfrac{3 \left(\ell^2 - m^2\right)}{\ell^2 + m^2}\right)^2 + \left(\dfrac{6 \ell m}{\ell^2 + m^2}\right)^2 + 24}$
i.e. $\;$ $r = \dfrac{\sqrt{9 \ell^4 + 9 m^4 - 18 \ell^2 m^2 + 36 \ell^2 m^2 + 24 \ell^4 + 24 m^4 + 48 \ell^2 m^2}}{\ell^2 + m^2}$
i.e. $\;$ $r = \dfrac{\sqrt{33 \ell^4 + 33 m^4 + 66 \ell^2 m^2}}{\ell^2 + m^2}$
i.e. $\;$ $r = \dfrac{\sqrt{33 \left(\ell^4 + 2 \ell^2 m^2 + m^4\right)}}{\ell^2 + m^2}$
i.e. $\;$ $r = \dfrac{\sqrt{33 \left(\ell^2 + m^2\right)^2}}{\ell^2 + m^2}$
i.e. $\;$ $r = \sqrt{33}$