Unizor - Geometry2D - Apollonius Problems - Points and Lines

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Generally speaking, all problems related to construction of a circle tangential to three other geometrical objects, points, straight lines or circles, is referred to as one of the Apollonius problems. Obviously, there is a history behind it. We will not talk about this history and, for those interested, the Web is your best source for details.
Here we will spend some time to address these problems using some old, previously addressed, and some new techniques, wherever they are applicable.

Apollonius Problems 1

Problem PPP
Construct a circle passing through three given points, A, B and C, not lying on the same line.

Solution
This problem had been addressed before. Here is a recap.
A locus of points equidistant from two given points is a line passing through a midpoint of a segment connecting these two points and perpendicular to it (prove it!).
If three points A, B and C are not lying on the same line, the midpoint perpendicular to AB intersects the midpoint perpendicular to BC (prove it!), and their intersection is equidistant from all three points - A, B and C.
Incidentally, the consequence of this is that a perpendicular to a midpoint of AC also passes through the same intersection point, so all three midpoint perpendiculars intersecting at one point - a center of a circle passing through three given points (prove it!).

Problem PPL
Construct a circle passing through two given points, A and B, and tangential to a straight line d, presuming that both points are located on the same side from the line.

Solution
Solution is based on the Tangent-Secant Power theorem that states that, if from a point P outside a circle drawn a tangent with a point of tangency T and a secant intersecting a circle at points A and B, then
PA·PB = PT² (prove it!).
Given PA and PB, we can find PT as an altitude towards a hypotenuse in a right triangle, where the base of this altitude divides a hypotenuse into segments equal to PA and PB.
Knowing that, in case AB is not parallel to line d, we find their intersection P and find PT, thereby determining a point of tangency of line d and a circle.
Once this point T is found, two segment bisectors for segments AB and AT would give a center of a circle.
If AB ∥ d, point T lies on intersection of d and segment AB's bisector.

Problem PLL
Construct a circle passing through a given point A and tangential to two straight lines p and q.

Solution
A solution can be based on a transformation of scaling.
Consider a general case of an angle formed by lines p and q with a vertex M and point A inside this angle.
Assume a circle tangential to lines p and q is constructed and it passes through point A.
Connect vertex M with point A and consider a scaling of this picture relative to center of scaling positioned at vertex M with any factor. Obviously, lines p and q will be transformed into themselves, an old circle tangential to these lines will be transformed into a new circle also tangential to the same lines, and point A on an old circle will be transformed into another point A' lying on line MA an on a new circle.
Based on this, a construction can be started with any circle tangential to lines p and q (its center is, obviously, on a bisector of an angle formed by these lines). Consider a line MA and its intersection with this circle at point A' (two points will lead to two solutions).
Now all we have to do is apply a scaling that transforms point A' into point A. This scaling will transform a circle we started with into the one needed.
A case of parallel lines p and q is trivial since we can easily determine the radius of a circle we want to construct - it is half the distance between these lines.

Problem LLL
Construct a circle tangential to three straight lines p, q and r.

Solution
Lines should not be all parallel. If two are, the case is simple since we know the radius of a circle. So, consider a general case of three lines forming a triangle.
In this general case we deal with a problem of inscribing a circle into a triangle. As we know, it's center lies on intersection of its angles' bisectors.