Question about reflection in geometry

Question

Let C be a circle with center O and L a line. First, prove that if L goes through O, then the mirror image of L (except O) is L itself (except O). Now we assume that L is not going through O. Prove that when a point P moves along L, then the mirror image P' moves along a circle passing through O. Prove that P' goes through the entire circle except O. Thus, the mirror image of L is a circle except a single point.

Answer 1

Set up an (x,y)-coordinate system in such a way that the center of the circle is the origin, i.e. the point (0,0), and the line L is parallel to the y-axis, so its equation is x = some constant, say x = c. The point P then is

(c,y).

The distance from the center to P is

√(c² + y²).

Therefore the distance from (0,0) to the "reflection" is

1/√(c² + y²)

So move from (0,0) in the direction of the point (c,y), to a point whose distance from (0,0) is that specified above. Every point on that line is of the form t•(c,y). The distance from (0,0) to t•(c,y) is

t√(c² + y²).

So we must have

1/√(c² + y²) = t√(c² + y²)

and therefore

t = 1/(c² + y²).

Hence the point P' is (c,y)/(c² + y²).

The problem then is to prove that that point moves along a circle passing through the origin. Notice that as y approaches infinity, this point does approach the origin. And when y = 0, then this point is (1/c²,0). And symmetry will suggest that the center of the circle is halfway between that point and the origin. So find the distance between the point identified above as P', and the point halfway between the origin and (1/c²,0). Observe that that distance does not change as y changes. Hence P' moves on a circle.