# Tangent Circles

A semicircle T has diameter AB=25 Point P lies on AB with AP=16 ans C is on the semicircle such that PC is perpendicular to AB. A circle w is drawn so taht it is tangent to segment PC, segment PB and T, what is the radius of w?

My try: connect A and C to get a right triangle, we get that the length of CP, we get the length of CP which is 12, i now am having trouble finding the radius, i tried creating triangle CPB but i am not sure if CB is collinear with the circle's center. Any help?

• Note that $CP^2=AP\cdot BP$. So, $CP=12$. – CY Aries Apr 23 '18 at 15:36

Try to draw a figure first. Let's call $O$ the middle of $AB$, $O'$ the center of the small circle, and $M$ the point where the small circle and the $T$ semicircle are tangent. If a circle is tangent to a line, it means that the radius is perpendicular to that line. Therefore, if two circles are tangent, it means that the two centers and the tangent point are all on the same line.

$$OM=R\\O'M=r\\OO'=R-r$$

The small center is tangent to $OB$ at a distance $r$ from $P$, and tangent to $PC$ at a distance $r$ from $P$. Witing Pythagora's theorem for the $OO'P$, you get $$(R-r)^2=r^2+(OP+r)^2$$ When you plug in the values, $R=12.5$ and $OP=3.5$, you get $(12.5-r)^2=r^2+(3.5+r)^2$

Let $r$ be the radius of $w$.

Let $M$ and $N$ be centres of $T$ and $w$ respectively. Denote the foot of perpendicular from $N$ to $AB$ by $K$.

Then $MK=25\div 2-9+r=3.5+r$, $NK=r$ and $MN=25\div2-r=12.5-r$.

By Pythagoras' Theorem,

\begin{align*} r^2+(3.5+r)^2&=(12.5-r)^2\\ 2r^2+7r+12.25&=156.25-25r+r^2\\ r^2+32r-144&=0\\ (r-4)(r+36)&=0\\ r&=4 \end{align*}

The trick is noticing that three points, the center of the larger circle, the center of the smaller circle and the point tangency are co-linear.

with the given information you will find a right triangle with sides $r,r+3.5$ and the hypothenuse of $12.5-r$

Upon using the Pythagorean theorem, you get $r=4.$