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For a given circle, prove that the lines of intersections by circles that pass through two given points converge at one point.

Mathematics Asked by Taxxi on November 19, 2021

The problem is from Kiselev’s Geometry exercise 410:

Given a circle O and two points A and B. Through these points, several
circles are drawn such that each of them intersects with or is tangent
to the circle O. Prove that the chords connecting the intersection
points of each of these circles, as well as the tangents at the points
of tangency with the circle O, intersect (when extended) at one point
lying on the extension of AB.

It is rather lengthy, so here is a rough picture of its claim:

enter image description here

The black circle is the given circle, and the intersection of two red circles are the given points. The problem claims that the three lines should intersect at one point.

The problem is then used to solve exercise 411, which becomes rather trivial:

Using the result of the previous problem, find a construction of the
circle passing through two given points and tangent to a given circle.

I could not do much about this exercise 410. I have found some solutions of exercise 411 which use circle inversion, but not only I am not familiar with the method (the book does not cover it), but also I am not sure if it also solves exercise 410. If the solution to the first problem requires inversion as well, I would very much appreciate if the basics of the method is also provided briefly as well.

One Answer

Let a circle through $A$ and $B$ cut the fixed circle (of radius $r$) at $F$ and $G$, and suppose lines $AB$, $FG$ intersect at $H$. If $D$ is the midpoint of $AB$, by power of a point we have: $$ HO^2-r^2=HFcdot HG=HAcdot HB=HD^2-AD^2, $$ that is: $$ HD^2-HO^2=AD^2-r^2. $$ As $AD$ and $r$ are given, that means $H$ belongs to the locus of points having a fixed difference of squared distances from two given points $O$ and $D$. It is well known (and easy to prove) that this locus is a line, perpendicular to line $OD$ and intersecting it at a point $E$. Hence $H$ lies on the intersection of this fixed line with line $AB$, whatever the radius of the variable circle.

enter image description here

Answered by Intelligenti pauca on November 19, 2021

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