Intersection of line with circle as function of distance from circle

Question

I'm trying to figure out if there is an analytical expression for this problem, or if it's going to be a transcendental equation no matter what:

Imagine that you have a circle of radius R centered at the origin (see figure below). There is a point directly above the circle at a distance D from the origin. As a function of D, is it possible to predict where a line from the point will intersect the circle?

In other words, knowing R and D, can I find X? (If I can find X, I can find Y, obviously).

I ran some numerical calculations, and you can see that as D gets large, X converges to R, which is exactly what you'd expect. And, as D approaches R (in this case I picked R = 1), X goes to zero.

I tried to fit this data with a function, and the best I can come up with with is:

x/R = ArcTan[aD^2 + bD + c]

It's an ugly function, but it works. Any help solving this analytically is appreciated!

Answer 1

Just use $$R^2=D\sqrt{R^2-x^2},$$ which gives $$x=\frac{R}{D}\sqrt{D^2-R^2}.$$ Indeed, let $A$ be a exterior point, $B$ be a touching point, $O$ be a center of the circle and $BC$ be a perpendicular to $AO$.

Thus, $\Delta ABO\sim\Delta BCO,$ which gives $$\frac{AO}{BO}=\frac{BO}{CO}$$ or $$BO^2=AO\cdot CO.$$

Answer 2

Suppose that point $P$ is an external point from which a tangent $PQ$ is drawn at point $Q$ to the circle of center $O$ & radius $R$ then drop a perpendicular $QS$ of length $x$ to the line $PO$ (where $PO=D$).

Now, in similar right triangles $\triangle QSO$ & $\triangle PQO$ $$\frac{QS}{PQ}=\frac{OQ}{PO}$$ $$\frac{x}{\sqrt{PO^2-OQ^2}}=\frac{R}{D}$$ $$\frac{x}{\sqrt{D^2-R^2}}=\frac{R}{D}$$ $$x=\frac{R}{D}\sqrt{D^2-R^2}$$