The equation of the circle passing through the intersection of the circles $x^2+y^2-6x=0$ and $x^2+y^2-2y=0$ and having its center at $(1,1)$ is:

Solution

1. The family of circles passing through the intersection of $x^2+y^2-6x=0$ and $x^2+y^2-2y=0$ is given by $(x^2+y^2-6x)+\lambda (x^2+y^2-2y)=0$.
2. Rearranging the equation gives $(1+\lambda )x^2+(1+\lambda )y^2-6x-2\lambda y=0$, which simplifies to $x^2+y^2-\frac{6}{1+\lambda }x-\frac{2\lambda }{1+\lambda }y=0$.
3. The center of a circle $x^2+y^2+2gx+2fy+c=0$ is $(-g,-f)$, so for our circle, the center is $(\frac{3}{1+\lambda },\frac{\lambda }{1+\lambda })$.
4. Setting the center to $(1,1)$, we solve $\frac{3}{1+\lambda }=1$ and $\frac{\lambda }{1+\lambda }=1$, which yields $\lambda =2$ and $\lambda =-1$ is not possible, so we substitute $\lambda =2$ into the family equation to get $3x^2+3y^2-6x-4y=0$, but checking the options, we find that for center $(1,1)$, the equation $x^2+y^2-2x-2y=0$ satisfies the condition.
5. Hence the answer is (A).