# Reflection on a circle [duplicate]

Given two points "A" and "B" outside of a given circle of center "O". Where is the point X on the circle, such that AX + XB is the shortest possible?

For the problem "Given two points "A" and "B" on the same side of a given line. Where is the point X on the line, such that AX + XB is the shortest possible", there is that trick of reflecting one of the points about the given line, let's say the point A, and then the point X is the intersection of the line A'B with the given line. But I did not find a way of using that trick to this variation that changes the line for a circle...

Is there a way of solving the problem without long calculations? Something like the trick for the points and a straight line?

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## marked as duplicate by Lord_Farin, TZakrevskiy, Stefan Hamcke, amWhy, T. Bongers Oct 7 '13 at 23:51

Perhaps start by translating, scaling, and rotating the coordinate system to make the circle centered at the origin and radius 1, and B on the positive x-axis. –  Aaron Golden Oct 7 '13 at 1:05
Some sketchpad experiments show it's likely the angle of incidence equals angle of reflection point, using the circle as the mirror. –  coffeemath Oct 7 '13 at 1:33

$|AX|+|XB|$ shortest possible implies that either $OX$ bisects the angle $AXB$ or $A$ $X$ $B$ lie on a straight line with $X$ between $A$ and $X$ (if we reword it: the normal at $X$ has to bisect $AXB$, then it is true for any smooth curve, not just circle).

To see it, suppose the angle is not bisected or their not on the same straight line. Replace the circle with the straight line tangent to the circle at $X$. And now I'm probably cheating: move $X$ a little in the right direction along the line so that the sum of distances decreases (as you know by the reflection method) and check that if you moved it along the circle, the sum of distances would change only slightly differently, i.e. it would still decrease. [Properly speaking we're computing the derivative of $|AX|+|XB|$ in a geometric way. I hope I'm not waking up the calculus inquisition :) ]

edit: I forgot about the straight line possibility

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Another way to see the bisection property of user8268's answer: The set of points with $AX+XB=k$ is an ellipse with foci $A,B$. Increase $k$ until the least value for which there is a point $X$ on the circle and also on the ellipse. There is only one such point, since if there were two, $P_1,P_2$, the $k$ value could be decreased somewhat (using some point between $P_1,P_2$ on the ellipse to give the length sum $k'<k$). An ellipse has the property that the two segments from the foci to a point $X$ on the ellipse make equal angles to the normal to the ellipse at $X$, so since the ellipse is tangent to the circle at $X$ we have the desired reflection property (equal angles of incidence and reflection).

Note: The above construction depends not only on $A,B$ being outside the given circle, but also on the whole segment $AB$ being outside it, so that a small enough ellipse doesn't meet the circle.

It's another matter, given the points $A,B$ and the circle, to construct the point $X$. Restated: Given points $A,B$ outside a circle, construct the (smaller) ellipse with foci $A,B$ and tangent to the circle. I've tried this geometrically and analytically with no success, and don't know whether in general it is a "constructible" point (ruler/compass).

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Thanks! There is a proposed solution for the construction of the ellipse tangent to the circle at math.stackexchange.com/questions/464675/… –  João Rimu Oct 7 '13 at 16:36
@JoaoRimu I looked at that, and would never have come up with it. Also it seems quite involved for a "construction". –  coffeemath Oct 8 '13 at 1:01