Answer
$r^2+4r \cos \theta-10r \sin \theta=-13$
Work Step by Step
Conversion of polar coordinates and Cartesian coordinates are as follows:
a) $r^2=x^2+y^2 \implies r=\sqrt {x^2+y^2}$
b) $\tan \theta =\dfrac{y}{x} \implies \theta =\tan^{-1} (\dfrac{y}{x})$
c) $x=r \cos \theta$
d) $y=r \sin \theta$
Since, we have $x=r \cos \theta$ and $y=r \sin \theta$ and $r^2=x^2+y^2 \implies r=\sqrt {x^2+y^2}$
Thus, we have an equivalent polar equation $(r \cos \theta+2)^2+(r \sin \theta-5)^2=16$ or, $r^2( \cos^2 \theta+ \sin^2 \theta)+4r\cos \theta-10r \sin \theta =16-4-25$
Thus, $r^2+4r \cos \theta-10r \sin \theta=-13$
