how would you go about using translated polar coordinates $y= rsin(theta) +5$ and $x=rcos(theta)$ and to evaluate double integrals with $f(x,y)= x^2+y^2$ bound by the circle $x^2+y^2 = 2x$? Please and thanks. I just need a general idea.
Based on original information (including first modification of question)
$x^2 + y^2 = r^2,\quad x = r\cos \theta,\quad y = r\sin\theta$
If bounded by $x^2 + y^2 = 2x$, that means the function is bounded by $$r^2 = 2r\cos \theta \iff r^2 - 2\cos\theta = 0 \iff r(r-2\cos\theta) = 0 \iff r = 0,\; r=2\cos \theta$$ So your bounds of integration with respect to $r$ are from $r = 0$ to $r = 2\cos \theta$
Hint: $x^2+y^2 = r^2$ and $dx\, dy = r\, dr\,d\theta$ (same is true for $dy\,dx$)
For a boundary of $x^2+y^2=2x$, transforming this into polar yields the equation $r^2=2r\cos\theta$. Solving this for $r$ gives two answers $r=0$ and $r=2\cos\theta$ (since
$r^2=2r\cos\theta \Longleftrightarrow r^2-2r\cos\theta = 0 \Longleftrightarrow r(r-2\cos\theta)=0$.
These solutions provide your limits for $r$.