Answer
(a) center $(\frac{a}{2}, \frac{b}{2})$, radius $\sqrt {\frac{a^2+b^2}{4}}$
(b) See graph.
Work Step by Step
(a) Convert the equation into polar coordinates using $r^2=x^2+y^2, x=r\cdot cos\theta, y=r\cdot sin\theta$, we have $a\cdot cos\theta+b\cdot sin\theta=\frac{ax}{r}+\frac{by}{r}=r$ which gives $r^2=ax+by$ or
$x^2+y^2-ax-by=0$. Try to make perfect squares $(x-\frac{a}{2})^2+(y-\frac{b}{2})^2=\frac{a^2+b^2}{4}$ which shows that it is a circle with a center at $(\frac{a}{2}, \frac{b}{2})$ and a radius of $\sqrt {\frac{a^2+b^2}{4}}$
(b) Given $r=2sin\theta+2cos\theta$, we have $a=2, b=2$, and from (a) we know that it is a circle centered at $(1,1)$ with a radius of $\sqrt {\frac{2^2+2^2}{4}}=\sqrt 2$ as shown in the graph.
