If we are given Polar equation $r = 2a\sin(\theta-\theta_{0})$ where $a,\theta_{0}$ are constants, we are asked to transform this equation to Cartesian equation.
My attempt :
$\sin(\theta-\theta_{0})=\sin(\theta)\cos(\theta_{0})-\sin(\theta_{0})\cos(\theta)$
$$r = 2a\sin(\theta-\theta_{0})$$ $$r = 2a(\sin(\theta)\cos(\theta_{0})-\sin(\theta_{0})\cos(\theta))$$ Now performing the substitution we get : $$ x=r\cos(\theta) , y = r\sin(\theta) \implies \cos(\theta)=\frac{x}{r},\space \sin(\theta)= \frac{y}{r}$$
Plugging back we get :
$$ r = 2a(\frac{y}{r}\cos(\theta_{0})-\sin(\theta_{0})\frac{x}{r})$$ $$ r = \frac{2a}{r}(y\cos(\theta_{0})-x\sin(\theta_{0}))/r$$ $$ r^{2} = 2a(y\cos(\theta_{0})-x\sin(\theta_{0})) $$ $$ x^{2}+y^{2} = 2ya\cos(\theta_{0})-2ax\sin(\theta_{0}) $$
$$ y^{2} - 2ya\cos(\theta_{0}) = -2ax\sin(\theta_{0}) -x^{2} $$
Edit:
After further manipulation suggested by J. W. Tanner I arrived at a solution :
$$ (x+asin(\theta_{0}))^{2} + (y-acos(\theta_{0}))^2 = a^{2} $$
which of course is just a circle with a center of $C(-asin(\theta_{0}),acos(\theta_{0}))$ , and radius $ r = a $ .