## Circling Around

Suppose that the circles in the figure below are tangent to one another. The radius of the circle centered at $A$ is one unit, and the radius of the circle centered at $B$ is two units. What are the radii of the congruent circles centered at $O$ and $P$?

Source: NCTM Problem to Ponder May 2011

SOLUTION

Since the two circles $O$ and $P$ are congruent, the centers of circles $A,B,C$ are collinear. The radius of circle $C$ equals 3. Let $r$ be the radius of circle $P\left (x,y\right )$.

Distance $AP=\sqrt{\left (x-0\right )^2+\left (y-5\right )^2}$
$1+r=\sqrt{\left (x-0\right )^2+\left (y-5\right )^2}$
$\left (1+r\right )^2=x^2+\left (y-5\right )^2\qquad\qquad\qquad\text{(1)}$

Distance $BP=\sqrt{\left (x-0\right )^2+\left (y-2\right )^2}$
$2+r=\sqrt{\left (x-0\right )^2+\left (y-2\right )^2}$
$\left (2+r\right )^2=x^2+\left (y-2\right )^2\qquad\qquad\qquad\text{(2)}$

Distance $CP=\sqrt{\left (x-0\right )^2+\left (y-3\right )^2}$
$3-r=\sqrt{\left (x-0\right )^2+\left (y-3\right )^2}$
$\left (3-r\right )^2=x^2+\left (y-3\right )^2\qquad\qquad\qquad\text{(3)}$

Multiply Eq. (1) by $-1$ and add to Eq. (2)
$-\left (1+r\right )^2=-x^2-\left (y-5\right )^2$
$\left (2+r\right )^2=x^2+\left (y-2\right )^2$
—————————————
$\left (2+r\right )^2-\left (1+r\right )^2=\left (y-2\right )^2-\left (y-5\right )^2$
$\left (2+r+1+r\right )\left (2+r-1-r\right )=\left (y-2+y-5\right )\left (y-2-y+5\right )$
$\left (3+2r\right )\left (1\right )=\left (2y-7\right )\left (3\right )$
$3+2r=6y-21\qquad\qquad\qquad\qquad\qquad\text{(4)}$

Multiply Eq. (2) by $-1$ and add to Eq. (3)
$-\left (2+r\right )^2=-x^2-\left (y-2\right )^2$
$\left (3-r\right )^2=x^2+\left (y-3\right )^2$
—————————————–
$\left (3-r\right )^2-\left (2+r\right )^2=\left (y-3\right )^2-\left (y-2\right )^2$
$\left (3-r+2+r\right )\left (3-r-2-r\right )=\left (y-3+y-2\right )\left (y-3-y+2\right )$
$\left (5\right )\left (1-2r\right )=\left (2y-5\right )\left (-1\right )$
$5-10r=-2y+5\qquad\qquad\qquad\qquad\text{(5)}$

Multiply Eq. (5) by $3$ and add to Eq. (4)
$3+2r=6y-21$
$15-30r=-6y+15$
——————————
$18-28r=-6$
$-28r=-6-18$
$-28r=-24$
$r=\frac{-24}{-28}$
$r=\frac{6}{7}$

Answer: $\frac{6}{7}$