Find real numbers $s$ and $t$ such that $\Gamma$ is the graph of $r = 2s \cos (\theta + t)$ The question: Let $\Gamma$ be a circle that passes through the origin. Show that we can find real numbers $s$ and $t$ such that $\Gamma$ is the graph of
$r = 2s \cos (\theta + t).$
I did this so far:
We can use the general circle equation, $$(x-h)^{2}+(y-k)^{2} = h^2 + k^2.$$
We take this equation and expand it: $$x^2 -hx-hx+h^2 + y^2 -ky-ky+k^2 = h^2 + k^2.$$
Simplify: $$x^2 -2hx + y^2 -2ky = 0.$$
Using what we know, we can get that: $$x = r \cos \theta$$ and $$y = r \sin \theta.$$
Plugging it in, we get: $$r^{2} \cos^{2} \theta - 2hr \cos \theta + r^{2} \sin^{2} \theta - 2kr \sin \theta = 0.$$
This looks like the Pythagorean identity, just "squared"! We will keep $r$ on one side, while we move the other over to the right: $$r(\sin^{2}\theta+\cos^{2}\theta)=2h\cos\theta+2k\cos\theta.$$
Which will give us: $$r = 2h\cos\theta+2k\sin\theta.$$
I'm not sure how to continue, I'm stuck at this point.

From
$$r=2(h \cos \theta + k \sin \theta)$$
$$r=2\sqrt{h^2+k^2}\left( \frac{h}{\sqrt{h^2+k^2}} \cos \theta + \frac{k}{\sqrt{h^2+k^2}}\sin \theta \right)$$
We can pick $s=\sqrt{h^2+k^2}$
and we can pick $t$ to satisfy $\cos t = \frac{h}{\sqrt{h^2+k^2}}$ and $\sin t = -\frac{k}{\sqrt{h^2+k^2}}$.
Remark: handle the case when $h=k=0$ separately.
