How to construct such a circle? Only straight edge and compasses are allowed.How could we draw if we were to draw the circle tangent to two straight lines and a circle.

  • 1
    $\begingroup$ Your title says 'circle tangent to two given circles and a straight line' and your last sentence says 'circle tangent to two straight lines and a circle'. Which is correct? You mean both? $\endgroup$
    – mathlove
    Jan 3, 2014 at 4:47
  • 1
    $\begingroup$ @mathlove it is essentially the same question. $\endgroup$
    – Igor Rivin
    Jan 3, 2014 at 4:56
  • $\begingroup$ @IgorRivin: Thanks for pointing it out. $\endgroup$
    – mathlove
    Jan 3, 2014 at 5:04
  • $\begingroup$ Hint: If the line contains a set of points that have x values on the boundary $(M_0_x + r, M_1_x + r)$ (where $M_0$, $M_1$ are midpoints of circle and $r$ is the radius), then there is no such circle. Can you construct boundaries for the y-values? $\endgroup$
    – cygorx
    Jan 3, 2014 at 5:16

3 Answers 3


Inversion is great here!

The most important trick is to note that if you want a circle tangent to three given circles, then depending on which of the 8 possibilities (there could be less) you want, you can appropriately enlarge or shrink the radii of all four circles such that their centres remain the same and they all remain tangent but one of the three circles becomes a point.

Now the problem reduces to finding a circle tangent to two given circles and a given point. Inverting at that point causes the desired circle to become a tangent line to two given circles, which is now easy.

Finally, for the original problem and others like it, you can invert almost anywhere to get the problem to be equivalent to the above solved problem.

  • 1
    $\begingroup$ I should add that inversion can be easily done using only compass and straightedge, and hence also this whole construction. $\endgroup$
    – user21820
    Jan 27, 2014 at 2:23
  • 1
    $\begingroup$ When I thought about this problem (many years ago) I came up with essentially the same solution with a slightly different order of operations: I would first enlarge the circles to make sure that two of them intersect. Then invert w.r.t. the point of intersection to have the same problem with two lines and a circle. Then shrink the circle to become a point (while parallel transporting the lines correspondingly). The rest is a simple homothety. Bisect the angle between the lines etc. $\endgroup$ Jul 14, 2019 at 16:59
  • $\begingroup$ @JyrkiLahtonen: Very nice! =) $\endgroup$
    – user21820
    Jul 15, 2019 at 14:05

As pointed out by user21820, this answer only covered the very special case where the two circles and tangents are kissing each other. i.e. this answer didn't really address the given question properly. However, since the answer is sort of cute. I'll just leave it here as is.

I'm going to only cover the part of the problem in the title. Namely,

$\hspace0.5in$ how to construct a circle which is tangent to a pair of circles
$\hspace0.5in$ and a line which are tangent among themselves

The other part of the problem in question body:

$\hspace0.5in$ how to construct a circle which is tangent to a circle and
$\hspace0.5in$ a pair of lines which are tangent among themselves

is relatively simple. My only hint is the center of the desired circle can be obtained by drawing two pair of parallel lines.

Part 1: Geometric relations among various components

Give a pair of circles and a line tangent among themselves. Let $A$ and $B$ be the centers of the two circles. Let $C$ be the point the two circle touching each other. Let $D$ and $E$ be the point the two circles touching the line. Their relative positions are illustrated in the figure below. kissing circles - helper conics

Given an arbitrary circle, if it is tangent to both circles, elementary geometry tells us its center will be lying on a hyperbola passing through $C$ with $A$ and $B$ as its foci. This is the pink curve in above figure.

Similarly, if a circle is tangent to the first circle and the line, its center will be lying on a parabola passing through $D$ and $B$. If one reflect $A$ w.r.t the line $DE$ to the point $A'$ and draw a line parallel to $DE$ (the blue dashed line in above diagram). The parabola will have this line as its directix and $A$ as its focus.

The same happens to the second circle. If a circle is tangent to the second circle and the line $DE$, its center will be lying on a parabola passing $A$ and $E$ and having $B$ as its focus.

These two parabolas are the two orange curves in above diagram. As shown in the diagram, these two parabolas and a branch of the hyperbola intersect at a common point $F$. This will be the center of the desired circle. We can construct the desired circle by dropping a perpendicular from $F$ to the line $DE$. Let $G$ be the intersection with the line, $FG$ will then be the radius of the desired circle.

Part 2: Ruler and Compass only construction

kissing circles - constructions

Start with these geometric facts, it is enough to derive a "ruler and compass only" geometric construction of $F$, $G$ and hence the desired circle. To simplify description of the recipe, I will assume people already know

$\hspace0.5in$ Given two line segments of length $L_1$ and $L_2$, how to construct
$\hspace0.5in$ a line segment with length proportional to $\sqrt{L_1 L_2}$

As illustrated by diagram above, if one construct a pair of points $P$ on line $AD$, $Q$ on line $BE$ such that

$$|DP| = 2\sqrt{|AD||DE|}\quad\text{ and }\quad |EQ| = 2\sqrt{|BE||DE|}$$

The line $PQ$ will intersect the line $DE$ at $G$. If one construct a circle centered at $D$ with radius $DG$ and let it intersect the line $AD$ at $R$. The line $RE$ will intersect the line $FG$ at some point $S$ $\color{blue}{^{[1]}}$. The center of desired circle $F$ will be simply the mid-point between $G$ and $S$.

Part 3: Why these work

To see why these work. let $R_A, R_B$ be the radii of the two circles. Let $x_A = |DG|$, $x_B = |GE|$ and $y = |FG|$. We have

  • $x_A^2 = 4R_A y$ because $F$ lies on the parabola passing through $D$ and $B$.
  • $x_B^2 = 4R_B y$ because $F$ lies on the parabola passing through $E$ and $A$.

These two together implies $x_A : x_B = \sqrt{R_A} : \sqrt{R_B}$. That's why the line $PQ$ intersect the line $DE$ at $G$. Furthermore, it is easy to see

$$x_A + x_B = \sqrt{(R_A + R_B)^2 - (R_A - R_B)^2} = 2\sqrt{R_A R_B}$$

Since $x_A + x_B = \sqrt{4 R_A y} + \sqrt{4 R_B y}$, we get

$$\sqrt{y} = \frac{\sqrt{R_A R_B}}{\sqrt{R_A}+\sqrt{R_B}} \quad\implies\quad \frac{2y}{x_A + x_B} = \frac{\sqrt{R_A R_B}}{(\sqrt{R_A}+\sqrt{R_B})^2} = \frac{x_A x_B}{(x_A + x_B)^2} $$ That's why when $RE$ intersect $FG$ at $S$, $F$ is the mid-point between $S$ and $G$.

Finally, above equation in $y$ can be rewritten in a more symmetric form

$$\frac{1}{\sqrt{y}} = \frac{1}{\sqrt{R_A}} + \frac{1}{\sqrt{R_B}} \quad\implies\quad \left(\frac{1}{R_A} + \frac{1}{R_B} + \frac{1}{y}\right)^2 = 2\left(\frac{1}{R_A^2} + \frac{1}{R_B^2} + \frac{1}{y^2}\right) $$ The relation on the R.H.S is a special case of the Descrates Theorem for the Apollonius's kissing circles problem. The wiki pages for these two topics and the references there will have more information about other construction methods.


  • $\color{blue}{[1]}$ Since $FG$ is perpendicular to $DE$, we can construct the line $FG$ before we know the exact position of $F$.
  • $\begingroup$ The question didn't say that the line was tangent to the two circles. $\endgroup$
    – user21820
    Jan 27, 2014 at 2:17

There was already a complete version answer to the OP's question with details and discussions on different cases in general here: Apollonius's Construction by Hubert Shutrick

One example created via Geogebra per the principles is as:tangent circles animation


Not the answer you're looking for? Browse other questions tagged or ask your own question.