Congruent circles $\Gamma_1$ and $\Gamma_2$ have radius $2012,$ and the center of $\Gamma_1$ lies on $\Gamma_2.$ Suppose that $\Gamma_1$ and $\Gamma_2$  intersect at $A$ and $B$. The line through $A$ perpendicular to $AB$ meets $\Gamma_1$ and $\Gamma_2$ again at $C$ and $D$, respectively. Find the length of $CD$.

[i]Author: Ray Li[/i]
1. Let the centers of the circles $\Gamma_1$ and $\Gamma_2$ be $O_1$ and $O_2$ respectively. Since $\Gamma_1$ and $\Gamma_2$ are congruent with radius $2012$, we have $O_1O_2 = 2012$ because $O_1$ lies on $\Gamma_2$.
2. The circles intersect at points $A$ and $B$. The line through $A$ perpendicular to $AB$ meets $\Gamma_1$ again at $C$ and $\Gamma_2$ again at $D$.
3. Since $O_1$ and $O_2$ are the centers of the circles, $O_1A = O_1B = O_2A = O_2B = 2012$.
4. The line $CD$ is perpendicular to $AB$ and passes through $A$. Therefore, $CD$ is the perpendicular bisector of $AB$.
5. Since $O_1$ and $O_2$ are the centers of the circles and $A$ and $B$ are points of intersection, $O_1O_2$ is also the perpendicular bisector of $AB$.
6. The distance $O_1O_2 = 2012$ is the radius of the circles, and $O_1$ lies on $\Gamma_2$, so $O_1O_2 = 2012$.
7. The line $CD$ is parallel to $O_1O_2$ and is twice the radius of the circles because it spans from one side of $\Gamma_1$ to the other side of $\Gamma_2$.
8. Therefore, the length of $CD$ is $2 \times 2012 = 4024$.

The final answer is $\boxed{4024}$.