Find the center and radius of the circle $\;$ $r^2 - 4r \left(\sqrt{3} \cos \theta + \sin \theta\right) + 7 = 0$
Equation of given circle: $\;$ $r^2 - 4r \left(\sqrt{3} \cos \theta + \sin \theta\right) + 7 = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $\left(R, \phi\right)$ $\;$ be the polar coordinates of the center and the radius be $= a$.
Then equation $(1)$ should be identical with
$r^2 - 2R r \cos \left(\theta - \phi\right) + R^2 - a^2 = 0$
i.e. $\;$ $r^2 - 2R r \left(\cos \theta \cos \phi + \sin \theta \sin \phi\right) + R^2 - a^2 = 0$
$\therefore \;$ $4r \times \sqrt{3} \cos \theta = 2 R r \cos \theta \cos \phi$
i.e. $\;$ $R \cos \phi = 2 \sqrt{3}$ $\;\;\; \cdots \; (3a)$;
$4 r \sin \theta = 2 R r \sin \theta \sin \phi$
i.e. $\;$ $R \sin \phi = 2$ $\;\;\; \cdots \; (3b)$
and $\;$ $R^2 - a^2 = 7$ $\;\;\; \cdots \; (3c)$
From equations $(3a)$ and $(3b)$
$\tan \phi = \dfrac{1}{\sqrt{3}}$
$\implies$ $\phi = \tan^{-1} \left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{\pi}{6}$ $\;\;\; \cdots \; (4a)$
and $\;$ $R^2 \cos^2 \phi = 12, \;\; R^2 \sin^2 \phi = 4$
i.e. $\;$ $R^2 \left(\sin^2 \phi + \cos^2 \phi\right) = R^2 = 12 + 4 = 16$
$\implies$ $R = 4$
Substituting the value of $R^2$ in equation $(3c)$ gives
$16 - a^2 = 7$
i.e. $\;$ $a^2 = 9$ $\implies$ $a = 3$ $\;\;\; \cdots \; (4b)$
$\therefore \;$ The center of the circle is $\;$ $\left(4, \dfrac{\pi}{6}\right)$ $\;$ and the radius is $\;$ $3$.