If $x=h+a\cos \theta$ and $y=k+a\sin \theta$ eliminate $\theta$

$\begin{array}{l}x-h=a\cos \theta \\ \text{Squaring on both sides}\\ {(x-h)}^2=a^2{\cos }^2\theta \to \left(1\right)\\ y-k=a\sin \theta \\ \text{Squaring on both sides}\\ {(y-k)}^2=a^2{\sin }^2\theta \to \left(2\right)\\ \text{Add equation} \left(1\right)\text{\hspace{0.17em}and equation\hspace{0.17em}(2)}\\ {(x-h)}^2+{(y-k)}^2=a^2({\cos }^2\theta +{\sin }^2\theta )\\ {(x-h)}^2+{(y-k)}^2=a^2\end{array}$