11. Find the equation of the circle whose center lies on the line $x-2 y=0$, and which is tangent to the circle $C_{1}: x^{2}+y^{2}-2 x-4 y+4=0$ at the point $(1,1)$.
11. Consider the point $(1,1)$ as the point circle $C_{2}:(x-1)^{2}+(y-1)^{2}=0$. Then the condition “tangent to circle $C_{1}$ at point $(1,1)$” can be converted to “passing through the intersection points of circles $C_{1}$ and $C_{2}$”. Therefore, the family of circles can be used to solve the problem.
Thus, by the problem statement, we can set the equation of the required circle as:
$(x-1)^{2}+(y-1)^{2}+\lambda\left(x^{2}+y^{2}-2 x-4 y+4\right)=0$. Simplifying, we get
$x^{2}+y^{2}-2 x-\frac{2(1+2 \lambda)}{1+\lambda} y+\frac{2(1+2 \lambda)}{1+\lambda}=0$.
The center of the circle represented by the above equation is $\left(1, \frac{1+2 \lambda}{1+\lambda}\right)$, and since the center lies on the line $x-2 y=0$,
we have, $1-2 \times \frac{1+2 \lambda}{1+\lambda}=0$,
solving for $\lambda$ gives $\lambda=-\frac{1}{3}$.
Substituting $\lambda=-\frac{1}{3}$ into (1), we get the equation of the required circle as $x^{2}+y^{2}-2 x-y+1=0$.