circleseuclidean-geometrygeometry
$\angle$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.
For the two equations you have, expand the squares on the left sides. You'll have $s^2$ and $r^2$ on the left side in both equations. Subtracting the two equations gives an equation that is linear in $s$ and $r$. Solve the linear equation for one variable and substitute into one of your original two equations to solve for the other variable. I believe that there are two solutions (by symmetry over the line through the known center and the known circumference point).
edit Let me redefine your variables, then go through some of the algebra. I'll continue to call the known center $(h,k)$ and the radius of that circle $r_1$ and the radius of the other circle $r_2$; I'll call the known point on the circumference of the other circle $(a,b)$, the desired point of tangency $(c,d)$, and the unknown center of the other circle $(m,n)$. The system of equations you had, written in my variables, is: $$\begin{align} (a-m)^2+(b-n)^2&=r_2^2 \\ (m-h)^2+(n-k)^2&=(r_1+r_2)^2 \end{align}$$ We want to solve for $(m,n)$. Expand the squares (use $(x+y)^2=x^2+2xy+y^2$) on the left sides of both equations: $$\begin{align} a^2-2am+m^2+b^2-2bn+n^2&=r_2^2 \\ m^2-2mh+h^2+n^2-2nk+k^2&=(r_1+r_2)^2 \end{align}$$ Subtract the two equations (I'm subtracting the first from the second): $$(m^2-2mh+h^2+n^2-2nk+k^2)-(a^2-2am+m^2+b^2-2bn+n^2)=(r_1+r_2)^2-r_2^2$$ $$m(2a-2h)+n(2b-2k)+(h^2+k^2-a^2-b^2)=r_1^2+2r_1r_2$$ I've grouped the terms on the left side into $m$ times some constant, $n$ times some constant, and a constant term (only $m$ and $n$ are variables here); I'll continue to group terms in this way. Solve for one of the variables ($m$): $$\begin{align} m(2a-2h) &=r_1^2+2r_1r_2-n(2b-2k)-(h^2+k^2-a^2-b^2) \\ &=(r_1^2+2r_1r_2-h^2-k^2+a^2+b^2)-n(2b-2k) \end{align}$$ $$m=\frac{(r_1^2+2r_1r_2-h^2-k^2+a^2+b^2)-n(2b-2k)}{2a-2h}$$ Now, substitute this expression for $m$ in one of the two original equations (I'll use the first): $$(a-\frac{(r_1^2+2r_1r_2-h^2-k^2+a^2+b^2)-n(2b-2k)}{2a-2h})^2+(b-n)^2=r_2^2$$ This is a (very messy) quadratic equation in $n$ (it can be written in the form $()n^2+()n+()=0$; $n$ is the only variable; the rest are constants). While it's possible to continue to solve by hand, the symbolic manipulation is messy (though I can fill in more detail if necessary). The two solutions for $n$ (note the $\pm$) that result are:
$n=\scriptstyle\frac{a^2 (b+k)-2 a h (b+k)+h^2 (b+k)+(b-k) \left(b^2-k^2+r_1^2+2 r_1r_2\right)\pm\sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left(a^2-2 a h+(b-k)^2+(h-r_1-2 r_2) (h+r_1+2 r_2)\right)}}{2 \left((a-h)^2+(b-k)^2\right)}$
The next step would be to take this, put it back into the expression for $m$ to get two values of $m$. Here are the results I get using Mathematica:
$(m,n)=$ $\begin{matrix}\scriptstyle{\left( \frac{1}{2 (a-h) \left((a-h)^2+(b-k)^2\right)} \left(\scriptstyle{\begin{align} a^4&-2 a^3 h+2 a h^3-h^2 \left(h^2+(b-k)^2\right)+a^2 (b-k)^2+(a-h)^2 r_1 \left(r_1+2 r_2\right)\\ &+b \sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left((a-h)^2+(b-k)^2-\left(r_1+2 r_2\right){}^2\right)}\\ &-k \sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left((a-h)^2+(b-k)^2-\left(r_1+2 r_2\right){}^2\right)} \end{align}}\right)\right.},\\ \quad\quad\scriptstyle{\left. \frac{\left((a-h)^2+(b-k)^2\right) (b+k)+(b-k) r_1 \left(r_1+2 r_2\right)-\sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left((a-h)^2+(b-k)^2-\left(r_1+2 r_2\right){}^2\right)}}{2 \left((a-h)^2+(b-k)^2\right)}\right)}\end{matrix}$
or $(m,n)=$ $\begin{matrix}\scriptstyle{\left( \frac{1}{2 (a-h) \left((a-h)^2+(b-k)^2\right)} \left(\scriptstyle{\begin{align} a^4&-2 a^3 h+2 a h^3-h^2 \left(h^2+(b-k)^2\right)+a^2 (b-k)^2+(a-h)^2 r_1 \left(r_1+2 r_2\right)\\ &-b \sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left((a-h)^2+(b-k)^2-\left(r_1+2 r_2\right){}^2\right)}\\ &+k \sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left((a-h)^2+(b-k)^2-\left(r_1+2 r_2\right){}^2\right)} \end{align}}\right)\right.},\\ \quad\quad\scriptstyle{\left. \frac{\left((a-h)^2+(b-k)^2\right) (b+k)+(b-k) r_1 \left(r_1+2 r_2\right)+\sqrt{-(a-h)^2 \left((a-h)^2+(b-k)^2-r_1^2\right) \left((a-h)^2+(b-k)^2-\left(r_1+2 r_2\right){}^2\right)}}{2 \left((a-h)^2+(b-k)^2\right)}\right)}\end{matrix}$
Since we know the two circles are tangent at the point $(c,d)$, the point $(c,d)$ is on the line connecting the centers of the circles and $\frac{r_1}{r_1+r_2}$ of the way from $(h,k)$ to $(m,n)$, so $$(c,d)=\left(\frac{r_2}{r_1+r_2}h+\frac{r_1}{r_1+r_2}m,\frac{r_2}{r_1+r_2}k+\frac{r_1}{r_1+r_2}n\right).$$
Mathematica isn't generating any nice simplification of that expression with $m$ and $n$ substituted in, so I'll stop here.
In reference to this image
and complementing other answers, the points on the green arc are represented by the equation $$ r = (r_1-r_2)\cos\theta-\sqrt{r_2^2-(r_1-r_2)^2\sin^2\theta},\qquad|\theta|\leq\arcsin\left(\frac{r_2}{r_1-r_2}\right), $$ while the points on the red arc are represented by the equation $$ r = (r_1-r_2)\cos\theta+\sqrt{r_2^2-(r_1-r_2)^2\sin^2\theta},\qquad|\theta|\leq\arcsin\left(\frac{r_2}{r_1-r_2}\right). $$ In particular, the points on the arc from $A$ to $B$ are represented by the second of previous equations with $$ \arctan\left(\frac{r_2}{r_1-r_2}\right)\leq\theta\leq\arcsin\left(\frac{r_2}{r_1-r_2}\right). $$
Best Answer
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