Find the center and radius of the circle $\;$ $r^2 - 2r \left(24 \cos \theta - 7 \sin \theta\right) + 141 = 0$
Equation of given circle: $\;$ $r^2 - 2r \left(24 \cos \theta - 7 \sin \theta\right) + 141 = 0$ $\;\;\; \cdots \; (1)$
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. $\;$ $r^2 - 2R r \left(\cos \theta \cos \phi + \sin \theta \sin \phi\right) + R^2 - a^2 = 0$ $\;\;\; \cdots \; (2)$
$\therefore \;$ $-4r \cos \theta = -2 R r \cos \theta \cos \phi$
i.e. $\;$ $R \cos \phi = 24$ $\;\;\; \cdots \; (3a)$;
$14 r \sin \theta = -2 R r \sin \theta \sin \phi$
i.e. $\;$ $R \sin \phi = -7$ $\;\;\; \cdots \; (3b)$
and $\;$ $R^2 - a^2 = 141$ $\;\;\; \cdots \; (3c)$
From equations $(3a)$ and $(3b)$
$\tan \phi = \dfrac{-7}{24}$
$\implies$ $\phi = \tan^{-1} \left(\dfrac{-7}{24}\right)$ $\;\;\; \cdots \; (4a)$
and $\;$ $R^2 \cos^2 \phi = 576, \;\; R^2 \sin^2 \phi = 49$
i.e. $\;$ $R^2 \left(\sin^2 \phi + \cos^2 \phi\right) = R^2 = 579 + 46 = 625$
$\implies$ $R = 25$
Substituting the value of $R^2$ in equation $(3c)$ gives
$625 - a^2 = 141$
i.e. $\;$ $a^2 = 484$ $\implies$ $a = 22$ $\;\;\; \cdots \; (4b)$
$\therefore \;$ The center of the circle is $\;$ $\left(25, \tan^{-1} \left(\dfrac{-7}{24}\right)\right)$ $\;$ and the radius is $\;$ $22$.