11. Given two circles $x^{2}+y^{2}-2 a x=0, x^{2}+y^{2}-2 b y=0$, where $b \neq 0$, the equation of the circle with the common chord as its diameter is $\qquad$ .
11. $\left(a^{2}+b^{2}\right)\left(x^{2}+y^{2}\right)-2 a b(b x+a y)=0$. Hint: Set the equation of the circle as $\left(x^{2}+y^{2}-2 a x\right)+k\left(x^{2}+y^{2}-2 b y\right)$ $=0$, i.e., $x^{2}+y^{2}-\frac{2 a}{k+1} x-\frac{2 b k}{k+1} y=0(k \neq-1),\left(x-\frac{a}{k+1}\right)^{2}+\left(y-\frac{b k}{k+1}\right)^{2}=A$ (here $A$ does not need to be written out), the center $O_{1}$ is $\left(\frac{a}{k+1}, \frac{b k}{k+1}\right)$, and the common chord equation is $a x=b y, O_{1}$ lies on it, thus $k=\frac{a^{2}}{b^{2}}$.