PMO: Circ

Let the center of the circle be $O$ , its radius be $r$ and it intersects $BC$ at $P$ . Since $AB$ is tangent to the circle at $A$ , $\angle OAB = 90^\circ$ . Also since $CP$ is a diameter of the circle, $\angle ADP = 90^\circ$ . As $CD || AB$ , $\angle DCB = \angle ABC= \theta$ . Therefore, $\triangle ABO$ is similar to $\triangle CDO$ .

In $\triangle ABC$ , by Pythagorean theorem,

$\begin{aligned} OB^2 & = OA^2 + AB^2 \\ (12-r)^2 & = r^2 + 6^2 \\ 144 - 24r + r^2 & = r^2 + 36 \\ 12 - 2r & = 3 \\ 2r & = 12 - 3 \\ \implies r & =4.5 \end{aligned}$

From the two similar triangles,

$\begin{aligned} \frac {CD}{CP} & = \frac {AB}{OB} \\ \frac {CD}{2r} & = \frac 6{12-r} \\ \implies CD & = \frac {12r}{12-r} = \frac {54}{7.5} = \boxed{7.2} \end{aligned}$