Find the radius of a circle touching three circles

Question

As an exercise for myself, I made a diagram where the goal is to solve for $w$ when $x$, $y$, $z$, and $k$ are known. I know this problem already exists but I'm having trouble locating it online:

The circle at $D$ is touching the circles at $A$, $B$, and $C$ at a single point. To find $w$, I decided I would calculate the area of quadrilateral $ABCD$ in two different ways: using triangles $ABD + BCD$ and $ACD + ABC$. I used Heron's formula 4 times:

$$\sqrt{(x+2y+z+k)(x+2y+z-k)(x-z+k)(z-x+k)}$$$$+$$$$\sqrt{(x+2w+z+k)(x+2w+z-k)(x-z+k)(z-x+k)}$$$$=$$$$4\sqrt{xyw(x+y+w)}+4\sqrt{zyw(z+y+w)}$$

Isolating $w$ is proving to be very difficult and I'm getting stuck. I was able to eliminate all the radicals, but the terms explode when I try to expand. Is there an easier way to find the radius of the circle at point $D$? Am I missing an obvious clue to make the problem easier?

Answer 1

Let us have a geometric vision of this issue.

Let us give the names $C_A,C_B,C_C$ for the initial circles, and $C_D$ for the unknown circle.

Let us consider separately the constraints on point $D$:

• circle $C_D$ must be tangent to $C_B$ and $C_A$ ; therefore its center must be at distance $d+b$ from $B$ and $d+a$ from $A$. Consider the difference of these distances (which eliminates unknown $d$) : this difference in absolute value is equal to $|b-a|$. But the set of points whose differences of distances to two fixed points is the geometrical definition of a hyperbola. As a consequence, point $D$ must belong to one of the two branches of a certain hyperbola with foci $A$ and $B$ (featured in green on the graphics), one can even say that it is the branch passing through tangency point $E$.

• for the same reason, $D$ must belong to one of the two branches of another hyperbola with foci $B$ and $C$ (featured in blue), the branch passing through tangency point $F$.

Therefore, $D$ is situated at the intersection of these two curves, provided they have a common intersection.

Here is a case with two solutions $D_1$ and $D_2$ (animated version here : points $A,B,C$ are movable, but must be kept in a certain range for the figure to be "reliable").

This qualitative approach can give, in a second step, a quantitative result i.e., provide explicit coordinates for the intersection point(s) through the resolution of a system of two quadratic equations, yielding a fourth degree equation either in $x$ or in $y$ which will be preferably solved by numerical means.

Answer 2

Here's a geometric ruler-and-compass construction. We want to find the circles tangent to three other circles (black in figure below), one of which touches the other two. This is a special case of Apollonius' Problem, which can be easily solved through inversion.

Let's draw a circle (dashed red) centred at one of the two tangency points and construct the inverse of the black circles with respect to the red one. The images of the circles touching at the centre of the inversion circle are two parallel lines (blue), passing through the intersection points of each circle with the red circle. The image of the third black circle is the small blue circle, tangent to one of the lines.

It is then easy to find two circles (orange) tangent to both lines and to the blue circle. Inverting back the orange circles with respect to the red one will give two circles (green) which are tangent to all three black circles.

Answer 3

I'm answering my own question with some algebra techniques (software) to find a solution to the equation with four radicals. I got the idea from the MathPages blog. This is a more general version of the problem, where the solution circle has radius $r$ and the other three circles have radii $a$, $b$, and $c$. The distances between the outer circle centers are $A$, $B$, and $C$:

The equation can be written as:

$$\sqrt{K}=\sqrt{X}+\sqrt{Y}+\sqrt{Z}$$

where: $$K=(A + B + C) (-A + B + C) (A - B + C) (A + B - C)$$ $$X=(A + 2 r + b + c) (-A + 2 r + b + c) (A - b + c) (A + b - c)$$ $$Y=(B + 2 r + a + c) (-B + 2 r + a + c) (B - a + c) (B + a - c)$$ $$Z=(C + 2 r + a + b) (-C + 2 r + a + b) (C - a + b) (C + a - b)$$

Clearing the radicals provides us with:

$$[2(X^2+Y^2+Z^2+K^2)-(X+Y+Z+K)^2]^2 - 64 X Y Z K=0$$

After substituting and expanding to get an enormous 8th degree polynomial (too large to write down here) of $r$, we can use algebra software to show that the following quadratic polynomial can be factored out:

$$r^2 \cdot (A^4 - 2 A^2 B^2 - 2 A^2 C^2 + 4 A^2 a^2 - 4 A^2 a b - 4 A^2 a c + 4 A^2 b c + B^4 - 2 B^2 C^2 - 4 B^2 a b + 4 B^2 a c + 4 B^2 b^2 - 4 B^2 b c + C^4 + 4 C^2 a b - 4 C^2 a c - 4 C^2 b c + 4 C^2 c^2)$$ $$+$$ $$r \cdot (2 A^4 a - 2 A^2 B^2 a - 2 A^2 B^2 b - 2 A^2 C^2 a - 2 A^2 C^2 c + 4 A^2 a^3 - 2 A^2 a^2 b - 2 A^2 a^2 c - 2 A^2 a b^2 - 2 A^2 a c^2 + 2 A^2 b^2 c + 2 A^2 b c^2 + 2 B^4 b - 2 B^2 C^2 b - 2 B^2 C^2 c - 2 B^2 a^2 b + 2 B^2 a^2 c - 2 B^2 a b^2 + 2 B^2 a c^2 + 4 B^2 b^3 - 2 B^2 b^2 c - 2 B^2 b c^2 + 2 C^4 c + 2 C^2 a^2 b - 2 C^2 a^2 c + 2 C^2 a b^2 - 2 C^2 a c^2 - 2 C^2 b^2 c - 2 C^2 b c^2 + 4 C^2 c^3)$$ $$+$$ $$A^4 a^2 + A^2 B^2 C^2 - A^2 B^2 a^2 - A^2 B^2 b^2 - A^2 C^2 a^2 - A^2 C^2 c^2 + A^2 a^4 - A^2 a^2 b^2 - A^2 a^2 c^2 + A^2 b^2 c^2 + B^4 b^2 - B^2 C^2 b^2 - B^2 C^2 c^2 - B^2 a^2 b^2 + B^2 a^2 c^2 + B^2 b^4 - B^2 b^2 c^2 + C^4 c^2 + C^2 a^2 b^2 - C^2 a^2 c^2 - C^2 b^2 c^2 + C^2 c^4$$

The solutions of this quadratic represent the possible radii of the solution circle. Desmos demo here. Here is a python program using SymPy to do the factoring:

from sympy import *
a, b, c, A, B, C, r = symbols('a b c A B C r')
K = (A + B + C) * (-A + B + C) * (A - B + C) * (A + B - C)
X = (A + 2 * r + b + c) * (-A + 2 * r + b + c) * (A - b + c) * (A + b - c)
Y = (B + 2 * r + a + c) * (-B + 2 * r + a + c) * (B - a + c) * (B + a - c)
Z = (C + 2 * r + a + b) * (-C + 2 * r + a + b) * (C - a + b) * (C + a - b)
expr = (2 * (X ** 2 + Y ** 2 + Z ** 2 + K ** 2) - (X + Y + Z + K) ** 2) ** 2 - 64 * X * Y * Z * K
print(expr.factor())