Unless you know the radius of the inner circle, there are infinitely many points which $D$ could be. Once you know the radius of the inner circle you will be able to find two possible points for $D$.
Supposing you know the length of $\overline{AB}$ (or equivalently, $\overline{AD}$). Then, using vector notation, you know what $\mathbf D \cdot \mathbf D$ is. Now $\mathbf D$ and $\mathbf {D-C}$ are perpendicular, so $\mathbf D\cdot (\mathbf D - \mathbf C)=0\implies \mathbf D\cdot \mathbf D=\mathbf D\cdot\mathbf C$ so if we write $\mathbf D=(x,y)$, $\mathbf C=(c_1,c_2)$ and $\overline{AD}=k$ then the points of $\mathbf D$ are given by the intersection of a line and a circle:
$x^2+y^2=k^2=xc_1+yc_2$
This will have two points of intersection (unless $k=0$ or the line is tangent to the circle). Obviously if $k=0$ then $\mathbf D$ is at the origin as well. Now if $k\ne0$ and $c_1\ne0$, we can solve the linear equation as follows:
$x=-\frac {c_2}{c_1} y+\frac {k^2} {c_1}$
Now you can substitute this in the quadratic equation:
$\left(-\frac{c_2}{c_1} y+\frac {k^2}{c_2}\right)^2+y^2=k^2$ Simplifying this will yield a quadratic equation for $y$ without $x$, so you may use the quadratic formula to solve for $y$. This is where you will get two values for $y$, which you can then plug back in to the earlier linear equation to solve for $x$.
If $c_1=0$, the linear equation would already give a specific value for $y$, so you would just substitute that value into the quadratic equation to find two values for $x$.