# Finding the circles passing through two points and touching a circle

Given two points and a circle, construct a/the circle through the two points and touching the given circle.

I came across this problem in History of Numerical Analysis by H. Goldstein. I spent some time on this. I have a method of constructing it using radical axis. I am wondering if there is a more elementary construction.

• when unsure always use circle inversion :) (here wrt. a circle with the center in one of the 2 point - after inversion the point goes to infinity, so you now need to find a line passing though a point (the image of the other point) and touching a circle (the image of the original circle). Then apply the inversion again, and the image of the line is the circle you're looking for. – user8268 Apr 11 '11 at 21:03
• @user8268: nice, but couldn't you translate this into more classical language by choosing the new circle in such a way that it intersects the original circle orthogonally, so you only have to invert the second point and construct a tangent to that? I haven't thought this through, though. – t.b. Apr 11 '11 at 21:10
• @Theo: that's certainly the sanest choice for the circle – user8268 Apr 11 '11 at 21:23
• user8268's idea enhanced by selecting mirror circle centered at one of the points and radius equals the length of the tangent works. the way i solved was to find the radical center of the circle with the system of coaxal circles through the points. just draw a circle centered on the perpendicular bisector of the given points and cutting the given circle. extend the points of intersection to meet line determined by the two points. gives you the radical center. draw tangents from there giving you two contact points. – abel Apr 12 '11 at 12:36
• @user8268: I elaborated your comment into an answer. I hope you like it. – t.b. Jul 11 '11 at 15:14

Since your proposed solution in the comments doesn't always work, let me give a somewhat more detailed description of the construction than the one outlined by user8268 (I'm assuming the given points are outside the given circle):

when unsure always use circle inversion :) (here wrt. a circle with the center in one of the 2 point - after inversion the point goes to infinity, so you now need to find a line passing though a point (the image of the other point) and touching a circle (the image of the original circle). Then apply the inversion again, and the image of the line is the circle you're looking for.

• Given: Two points $\color{blue}{A},\color{blue}{B}$ and a circle $\color{blue}{c}$ with center $\color{blue}{C}$.
• Wanted: The two points $\color{red}{P,Q}$ on $\color{blue}{c}$ such that the circles through $\color{blue}{A},\color{blue}{B},\color{red}{P}$ and $\color{blue}{A},\color{blue}{B},\color{red}{Q}$ are tangent to $\color{blue}{c}$ (drawing a circle through three points is trivial).

Construction:

The idea is outlined in user8268's comment: Draw a circle $\color{lime}{e}$ around $\color{blue}{A}$ intersecting $\color{blue}{c}$ orthogonally. The point is that $\color{blue}{c}$ is invariant under circle reflection at $\color{lime}{e}$. The point $\color{blue}{A}$ is sent to infinity and the sought circles will therefore become tangents to $\color{blue}{c}$. Since the circles pass through $\color{blue}{B}$, the tangents must pass through the reflection $\color{blue}{B'}$ of $\color{blue}{B}$ at $\color{lime}{e}$. Thus here's the construction in brief:

1. Construct the circle $\color{lime}{e}$ around $\color{blue}{A}$ intersecting $\color{blue}{c}$ orthogonally.
2. Reflect $\color{blue}{B}$ at $\color{lime}{e}$ to get $\color{blue}{B'}$.
3. Find the tangents from $\color{blue}{B'}$ to $\color{lime}{e}$.
4. and 5: reflect the tangents at $\color{lime}{e}$ to find the circles.

So here we go in more detail:

1. Draw the circle $\color{lime}{e}$ with center $\color{blue}{A}$ intersecting the circle $\color{blue}{c}$ orthogonally: First draw the circle with diameter $\color{blue}{AC}$ (dashed circle below) and intersect it with $\color{blue}{c}$, then draw the circle $\color{lime}{e}$ with center $\color{blue}{A}$ through those intersection points. The radii of $\color{blue}{c}$ and $\color{lime}{e}$ will meet perpendicularly at the points of intersection by Thales, as $\color{blue}{AC}$ is the diameter of the dashed circle.

2. Reflect $\color{blue}{B}$ at the circle $\color{lime}{e}$ to obtain $\color{blue}{B'}$. Note that $\color{blue}{B'}$ must lie outside $\color{blue}{c}$ as $\color{blue}{B}$ lies outside $\color{blue}{c}$. Recall the construction if $\color{blue}{B}$ lies outside $\color{lime}{e}$: the point of intersection of the segment $\color{blue}{AB}$ with the segment through the points of intersection of $\color{lime}{e}$ with the circle through $\color{blue}{A}$ and $\color{blue}{B}$ around their midpoint: Exercise: Give the construction if $\color{blue}{B}$ lies inside $\color{lime}{e}$.

3. Find the points of contact $P'$ and $Q'$ of the tangents from $\color{blue}{B'}$ to $\color{blue}{c}$: Draw the circle $\color{orange}{d}$ with diameter $\color{blue}{B'C}$ and call $P'$ and $Q'$ the points of intersection of $\color{orange}{d}$ with $\color{blue}{c}$ (see step 1.):

4. Since reflection at $\color{lime}{e}$ swaps $A$ and infinity and leaves the circle $\color{blue}{c}$ invariant, the tangents from $\color{blue}{B'}$ to $\color{blue}{c}$ will become circles tangent to $\color{blue}{c}$ upon reflection at $\color{lime}{e}$. Therefore the points $\color{red}{P}$ and $\color{red}{Q}$ are the reflections of $P'$ and $Q'$ at $\color{lime}{e}$. Since $\color{blue}{c}$ is orthogonal to $\color{lime}{e}$ they are simply the second points of intersection of $\color{blue}{c}$ with the lines $\color{blue}{A}P'$ and $\color{blue}{A}Q'$ (since those lines and the circle $\color{blue}{c}$ are invariant under reflection at $\color{lime}{e}$):

5. Finally, draw the circles through $\color{blue}{A},\color{blue}{B},\color{red}{P}$ and $\color{blue}{A},\color{blue}{B},\color{red}{Q}$: voilà!

Exercise: Treat the case in which both points $\color{blue}{A}$ and $\color{blue}{B}$ are inside the circle $\color{blue}{c}$.

Hint: Reflect at the circle with center $\color{blue}{B}$ through $\color{blue}{A}$.

Added: In retrospect I like the second solution using reflection at the circle around $\color{blue}{B}$ through $\color{blue}{A}$ better than the one I gave, because it works in all cases that have a solution—either $\color{blue}{A}$ and $\color{blue}{B}$ both inside or both outside of the circle $\color{blue}{c}$. At some point I might add a detailed explanation of that as a second answer.

• Pictures created using GeoGebra. – t.b. Jul 10 '11 at 18:14
• Excellent figures and construction! btw, why not just say circle with BC' as diameter, etc? It is a bit confusing (at least to me) when you talk of midpoint etc. Also a bit more explanation would make this easier to understand. – Aryabhata Jul 10 '11 at 21:46
• @Aryabhata: I added a picture (hopefully) illustrating the case in which both points are inside the circle. – t.b. Jul 11 '11 at 15:17
• @Aryabhata: I added a second answer with some more pictures – t.b. Jul 16 '11 at 8:05
• Quite impressive. Definitely deserving of inversive-geometry :-) – robjohn Feb 6 '12 at 16:54

Here are some pictures illustrating the argument I alluded to at the end of my first answer. It has the virtue of working without distinguishing the cases whether the two points $\color{blue}{A}$ and $\color{blue}{B}$ are inside or outside the given circle $\color{blue}{c}$.

• Given: Two points $\color{blue}{A},\color{blue}{B}$ and a circle $\color{blue}{c}$ with center $\color{blue}{C}$.
• Wanted: The two points $\color{red}{P,Q}$ on $\color{blue}{c}$ such that the circles through $\color{blue}{A},\color{blue}{B},\color{red}{P}$ and $\color{blue}{A},\color{blue}{B},\color{red}{Q}$ are tangent to $\color{blue}{c}$ (drawing a circle through three points is trivial).

Before delving into the solution let's get rid of some degenerate cases:

Warm-up Exercise: Treat all the cases in which $\color{blue}{A} = \color{blue}{B}$ or one of $\color{blue}{A}$ or $\color{blue}{B}$ lies on $\color{blue}{c}$.

So, from this point on, we assume $\color{blue}{A} \neq \color{blue}{B}$ and that both lie either inside or outside the circle $\color{blue}{c}$.

The idea is the same as in the other answer (i.e. user8268's solution). Reflecting the configuration at the circle $\color{green}{d}$ with center $\color{blue}{B}$ through $\color{blue}{A}$ fixes $\color{blue}{A}$ and sends $\color{blue}{B}$ to infinity. It transforms the circle $\color{blue}{c}$ into a circle $\color{blue}{c'}$ and transforms the circles we're looking for into tangents from $\color{blue}{A}$ to $\color{blue}{c'}$ because $\color{blue}{B}$ is sent to infinity and circle reflection preserves angles. Finding the tangents from the point $\color{blue}{A}$ to the circle $\color{blue}{c'}$ is easy and we need only reflect the points of tangency $P',Q'$ back to $\color{blue}{c}$ to find $\color{red}{P}$ and $\color{red}{Q}$. Drawing the circles through $\color{blue}{A}, \color{blue}{B}, \color{red}{P}$ and $\color{blue}{A}, \color{blue}{B}, \color{red}{Q}$ is again straightforward.

So here's the solution in somewhat more detail:

1. Reflect the circle $\color{blue}{c}$ at the circle $\color{green}{d}$ through $\color{blue}{A}$: To find $\color{blue}{c'}$, draw the line through $\color{blue}{B}$ and $\color{blue}{C}$ (if $\color{blue}{B} = \color{blue}{C}$ draw an arbitrary line through $\color{blue}{B}$) and reflect its points of intersection with $\color{blue}{c}$ at $\color{green}{d}$. Then the circle $\color{blue}{c'}$ is the circle with diameter those two reflected points. (See also point 2. in my other answer for more details.)

Exercise: Prove that $\color{blue}{A}$ is always outside the circle $\color{blue}{c'}$.

2. Find the tangents from $\color{blue}{A}$ to $\color{blue}{c'}$: Call the points of tangency $P'$ and $Q'$.

3. Reflect the points $P'$ and $Q'$ at $\color{green}{d}$ to find $\color{red}{P}$ and $\color{red}{Q}$:

4. Draw the circles through $\color{blue}{A}, \color{blue}{B}, \color{red}{P}$ and $\color{blue}{A}, \color{blue}{B}, \color{red}{Q}$: done.

Remark: Note that the red circles pass through the second points of intersection of $\color{green}{d}$ with the tangents from $\color{blue}{A}$ to $\color{blue}{c'}$. This means that the construction admits a slightly simpler variant (omitting steps 3 and 4) if one doesn't care about the points $\color{red}{P}$ and $\color{red}{Q}$. I chose to explain it the way I did, as the more efficient method in this remark doesn't exhibit as clearly why it works and in order to see that, one needs to consider the points $\color{red}{P}$ and $\color{red}{Q}$ anyway.

As a final picture, the case that both $\color{blue}{A}$ and $\color{blue}{B}$ lie outside of $\color{blue}{c}$: