construct circle tangent to two given circles and a straight line [closed]

Solution 1:

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.

Solution 2:

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.

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

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.

Notes

• $\color{blue}{[1]}$ Since $FG$ is perpendicular to $DE$, we can construct the line $FG$ before we know the exact position of $F$.

Solution 3:

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: