$The equation of the circle pas\sin g through the point (2+5\cos \theta ,-3+5\sin \theta ) is$

$x=2+5\cos \theta , y=-3+5\sin \theta$

$\frac{(x-2)}{5}=\cos \theta \to \left(1\right)$

$\frac{(y+3)}{5}=\sin \theta \to \left(2\right)$

$Square the Equations and add {\left(1\right)}^2+{\left(2\right)}^2$

$\Rightarrow {\left(\frac{x-2}{5}\right)}^2+{\left(\frac{y+3}{5}\right)}^2={\cos }^2\theta +{\sin }^2\theta$

$=x^2+y^2-4x+6y-12=0$