Arbitrary circles equal to polar coordinates $\alpha$ is an arbitrary (random) circle that passes through the origin. 
Show that there are real numbers $s$ and $t$ such that $\alpha$ is the graph of $r = 2s \cos (\theta + t)$.
I believe that $r = 2s \cos (\theta +t)$ can be simplified down to $r = x^2+y^2$ using Cartesian coordinates. What next?
 A: The term $2s$ is responsible for the size of the circle. Generally, $r=Rcos\theta$ is a circle that passes through O. (I assume that this piece of information can be used) The letter $t$ in your equation is responsible for a "phaseshift", in other words, it will make the circle "start" at a different point, but it will not change the shape of the circle.
A: I might be way off base, but here is a bunch of stuff you can do. I'll first convert to Cartesian coordinates using the identity $\cos(\theta+t) = \cos(\theta)\cos(t)-\sin(\theta)\sin(t)$ and $x = r\cos(\theta),\space y = r\sin(\theta)$ to get $$x^2+y^2= r^2 \\ = 2sr\cos(\theta+t) \\ = 2s[r\cos(\theta)\cos(t)-r\sin(\theta)\sin(t)] \\ = 2s[x\cos(t)-y\sin(t)]$$ If we rearrange the equation $x^2+y^2 =  2s[x\cos(t)-y\sin(t)]$ we'll get $$y^2+y\cdot(2s\sin(t))+[x^2-2sx\cos(t)]=0$$ which is a quadratic in terms of $y$. Hence $$y = \frac{-2s\sin(t)\pm \sqrt{4s^2\sin^2(t)-4(x^2-2sx\cos(t))}}{2} \\ = -s\sin(t)\pm \frac{\sqrt{4s^2\sin^2(t)+8sx\cos(t)-4x^2}}{2}$$ Although the the above equation for $y$ looks nasty, you can verify that $y(0)=0$ if you take the positive solution of the root, so $y$ does define a curve that passes through the origin. Further, $$y+s\sin(t) = \pm \frac{\sqrt{4s^2\sin^2(t)+8sx\cos(t)-4x^2}}{2} \\ \implies (y+s\sin(t))^2 = s^2\sin^2(t)+2sx\cos(t)-x^2 \\=s^2(1-\cos^2(t))+2sx\cos(t)-x^2 \\ = -(x^2-2sx\cos(t)+s^2\cos^2(t))+s^2 \\ = -(x-s\cos(t))^2+s^2$$ hence, $$(y+s\sin(t))^2+ (x-s\cos(t))^2=s^2$$ which is precisely the formula of a circle in the Cartesian plane with radius $s$ centered at $(s\cos(t),-s\sin(t))$ and radius $s$. Hopefully some of this is useful?
