Find the radius of circle whose centre is point of contact of two circles

Question

$\angle$AOB is a given angle. Two circles of radius $r_1$ and $r_2$ touch $OA$, $OB$ and touch each other. Find the radius of another circle touching the sides of $AOB$ and having its center at the point of contact of the two given circles

I made the figure and thought it was not possible to express it just in terms of $r_1$ and $r_2$. Please tell me if I am right or wrong.

Answer 1

Let $\angle$ OC2A be $\bf{\theta}$

In the figure perpendicular from each center of circle is draw to its tangent,and it's distance will be equal to its radius

Radius of C1 = $r_1$ Radius of C2 = $r_2$ Let radius of $C_1$ = $R$

Now, \begin{align} sin(\theta) = \frac{r_1}{OC1} = \frac{R}{OC1 + r_1} = \frac{r_2}{OC1+r_1+r_2}\\ \\ \text{On solving these we get } \ OC1=\frac{r_1}{sin(\theta)};\ sin(\theta) = \frac{r_2-r_1}{r_2+r_1}\\ \text{Substituting and solving for R we get, }\ \fbox{R = $\frac{2r_1r_2}{r_1+r_2}$} \end{align}

GeoGebra Figure : Figure 1.1

Answer 2

Comment: How about this construction; the required radius is equal to the diameter of smaller circle. In the case your picture shows, where the centers of all circles are on bisector o angle AOB, you can use similarities to find $r$ in terms of $r_1$ and $r_2$.

For the case where all centers are on the bisector, $R=4 r_4$, where $r_4$ is the radius of small circle tangent to one ray of angle and two circles with radius $r_1$ and $r_2$. $r_4$ can be found using Descartes theorem:

$(\frac 1{r_1}+\frac 1{r_2}+\frac 1{r_3}+\frac 1{r_4})^2=2(\frac1{r_1^2}+\frac 1{r_2^2}+\frac 1{r_3^2}+\frac 1{r_4^2})$

Here $r_3=\infty$ which represent the line(ray of angle), so the equation reduces to:

$(\frac 1{r_1}+\frac 1{r_2}+\frac 1{r_4})^2=2(\frac1{r_1^2}+\frac 1{r_2^2}+\frac 1{r_4^2})$