# 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.

$$r=2(h \cos \theta + k \sin \theta)$$
$$r=2\sqrt{h^2+k^2}\left( \frac{h}{\sqrt{h^2+k^2}} \cos \theta + \frac{k}{\sqrt{h^2+k^2}}\sin \theta \right)$$
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.