In this write-up, I pose a solution to the following problem:
Given a line and a circle with center K. Take an arbitrary point P on the circle. Construct two circles tangent to the given circle at P and tangent to the line.
First, I supply the steps of a construction of the given problem. Then, I give a proof that the construction works.
I start with the given: a line and a circle with center K.
Then I choose a random point, P, on the circle.
If the two new circles are to be tangent to the given circle at point P, then the two new circles must be tangent to the line that is tangent to the given circle at point P. Therefore, I construct the line through center K and point P, because I know that radius KP is perpendicular to the tangent line that I need.
Next I construct the tangent line that is perpendicular to radius KP at point P. I also construct the intersection of this tangent line and the given line, and I call this point Q.
Since each of the two new circles must be tangent to line PQ at point P, then it follows that their centers must lie on line KP (since a radius of each circle must be perpendicular to line PQ at point P). So where on line KP must these centers lie?
Well, since the new circles must also be tangent to the given line, we know that each must also have a radius that is perpendicular to the given line. Therefore, the center of each circle must be equidistant from the given line and from point P.
My conjecture is to construct a pair of angle bisectors. Choose two points on the given line, one to the left of Q and one to the right of Q. Let them be points R and S. Construct the angle bisectors of PQR and PQS.
Construct the point of intersection of each angle bisector with the line KP. Call these points L and M. These points are the centers of the two new circles.
Finally, construct one of the circles using L as the center and P as a point on the circle, and construct the other circle using M as the center and P as a point on the circle.
To test this construction, click here for a GSP sketch to manipulate.
What I need to prove is that points L and M are really centers of the tangent circles.
To do this, I drop the perpendicular from L to the given line and call the point of intersection point T. I also drop the perpendicular from M to the given line and call the point of intersection point U.
Now, by construction, angles LQT and LQP are congruent. Also, angles LTQ and LPQ are right by construction and are thereby congruent. Since the sum of the angles of a triangle equal 180 degrees, then angles QLT and QLP are congruent. Since segment LQ is reflexive, we have that triangles LQT and LQP are congruent by ASA congruency.
Similarly, triangles MQP and MQU are congruent by ASA congruency.
Since corresponding parts of congruent triangles are congruent, segments LT and LP are congruent. Similarly, segments MP and MU are congruent.
It follows that L is the center of a circle, with points T and P being on that circle. Also, M is the center of a circle, with points P and U being on that circle.
Because LP and MP are perpendicular to line PQ, and LT and MU are perpendicular to the given line, then the two constructed circles are tangent to the given circle at point P and are tangent to the given line, as required. QED.