Given 3 circles:

  • $C_1$ centered at $(0,0)$ with radius 1
  • $C_2$ centered at $(a,0)$ with radius $a+1$
  • $C_3$ centered at $(-a,0)$ with radius $a+1$

(so $C_1$ is internaly tangent to both $C_2$ and $C_3$ )


What are the centres of circle $C_4$ and $C_5$ that are tangent to all three circles?

I know they are on the $x=0$ axis but what is their $y$ coordinate?

  • $\begingroup$ For $a=3$ we have $O_{4,5}=(0,\pm1.6).$ $\endgroup$ – Lucian Dec 14 '15 at 7:51

WOLOG, we will assume $C_4$ / $C_5$ are lying in upper / lower half-plane respectively.

Let $c_i$ and $r_i$ be the center and radius for circle $C_i$ where $i = 1,\ldots,5$.
By symmetry, $r_4 = r_5$. Let us call this common value $r$. In term of $r$, $$c_4 = (0,1+r)\quad\text{ and }\quad c_5 = (0,-1-r)$$

Since $C_4$ is inside and tangent to $C_2$, we have

$$|c_4 - c_2| = r_2 - r_4 \implies (r+1)^2 + a^2 = (a+1-r)^2 \implies r = \frac{a}{a+2} $$ This leads to $c_4 = -c_5 = \left( 0, 2\left(\frac{a+1}{a+2}\right)\right)$.


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