$CD$ is the common chord of the two circles of equal radii touching a line $L$ at $A$ and $B$. Let $C$ be closer to the line $L$ than $D$. The ratio of the radii of circumcircles of the triangle $AC$ and $ADB$ is ___.
Step-by-Step Solution
Key Concept: The angle relationships in the two circles combined with the sine rule for circumradius determine their ratio.
Let $R_1$ and $R_2$ be the circumradii of triangles $ACB$ and $ADB$ respectively. Using $R = \frac{AB}{2\sin(\angle)}$, we have $R_1 = \frac{AB}{2\sin(\angle ACB)}$ and $R_2 = \frac{AB}{2\sin(\angle ADB)}$. Since the common chord $CD$ bisects the common tangent and $\angle ACB = 180° - 2\theta$ while $\angle ADB = 2\theta$, we find $\frac{R_1}{R_2} = 1$.
Correct Answer: 1