For simplicity, and WOLOG (I like that acronym, thanks Bractals), let's assume the given circle of radius a is located on the line y = 0, and the center of the inversion circle of radius k is located at the origin, that is , coordinate (0,0). Then, we can ignore the y coordinates altogether.

Let's start by fixing the given circle's center at a distance c from the origin. The inverted image has a center located at c' = ck²/(c² -a²), where a is the radius of the given circle and k is the inversion radius.

Next, the distance between the line and the center of the given circle is b. The point of interest, L, on the line is located at x-coordinate (c-b). The inverted image of this point, L', is located at k²/(c-b), which is the point on the resulting circle farthest from the origin. The diameter of this circle extends from the origin to this point. The center of this circle is half of this distance, and so is located at k²/2(c-b).

The inversion images of the given circle and the given line are concentric circles, so their centers are located at the same point. Therefore, ck²/(c²-a²) = k²/2(c-b). This simplifies to c² -2bc + a² = 0. Using the quadratic formula, we get c = b +- (b² - a²)^.5.

Now, let's determine the location of the given point by starting from the location of its inversion image at p' = k²/2(c-b), which is the center of the concentric circles we obtained from inversion of the given line and circle. Since |p|*|p'| = k², we have p = 2(c-b).

The distance from p to the line is |p - L|, which gives us the equation |p-L| = |2(c-b) - (c-b)| = |c-b|. Now we can substitute for c. So, |p-L| = |[b+- (b² - a²)^.5] - b|.

Our solution, then, is |p-L| = (b² - a²)^.5.

Edited to correct typos in key equations as pointed out by Eric. Thanks, Eric.