This solution definitely seems to have problem(s), but perhaps even though it's wrong, it'll help get someone to a complete/correct solution.
First, suppose that the point is outside the circle. For each secant from the point through the circle, the distance from the given point to the midpoint of the part of the secant that is inside the circle is equal to the average of the distances to all of the points in the circle through which that secant passes. The locus of these midpoints is an arc of the circle that has a diameter with endpoints at the given point and the center of the given circle.
If we place the given point at the origin and the center of the given circle at $d$ on the positive horizontal axis, the locus of points described above has polar equation $r=\frac{d}{2}\cos\theta$ for $-\arcsin\frac{r}{d}\le\theta\le\arcsin\frac{r}{d}$. For each $\theta$, the length of the line segment on the secant and inside the circle is $2\sqrt{r^2-d^2\sin^2\theta}$.
So, the average distance should be: $\frac{1}{2\arcsin\frac{r}{d}}\int_{-\arcsin\frac{r}{d}}^{\arcsin\frac{r}{d}}\left(2\sqrt{r^2-d^2\sin^2\theta}\cdot\frac{d}{2}\cos\theta\right)d\theta,$ which I let Mathematica work on for a bit and it's telling me is 0, so I probably screwed something up.
Now, if the point is inside the circle, then we have the entirety of the circle with diameter with endpoints at the given point and the center of the circle. Placing the given point and the circle as above, the locus-circle has the same equation, $r=\frac{d}{2}\cos\theta$, the length of the secant line segments inside the circle (now just chords) is the same $2\sqrt{r^2-d^2\sin^2\theta}$, but the limits of integration change to encompass the whole circle: $\frac{1}{\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(2\sqrt{r^2-d^2\sin^2\theta}\cdot\frac{d}{2}\cos\theta\right)d\theta.$