The circles $C_{1},C_{2},C_{3}$ with radii $1,2,3$ respectively,touch each other externally. The centres of $C_{1}$ and $C_{2}$ lie on the x-axis ,while $C_{3}$ touches them from the top. Find the ordinate of the centre of the circle that lies in the region enclosed by the circles $C_{1},C_{2},C_{3}$ and touches all of them.
Okay, I can see the lines joining the centres of the circle form a right angled triangle with sides $3,4,5$. But, I can't prosper furthur..any hint or solution.
On
This is not an answer, but I wanted to upload a picture to make the problem more clear to everyone.
@Viku, is this your problem? Finding the $y$ coordinate of the center of the bright red circle?
And this is a triangle you mentioned connecting the centers of $C_1,C_2,C_3$.
And the close-up of the enclosed region and the inner cirlce.
The $y$ coordinate of the center appears to be around $0.87$.
On
HINT...if $(a,b)$ is the centre of the circle and its radius is $r$ you can set up and solve a system of three simultaneous equations. So for example, for circle $C_1$ you have $$(a+1)^2+b^2=(r+1)^2$$ and likewise for the other two circles.
Of course, quoting Descartes' Theorem will be a short-cut to finding $r$ but you will still need two of these equations to find $(a,b)$
The expressions are bit complicated, I'll state some facts of outer and inner Soddy circles without proofs.
Denote the corresponding quantities for outer and inner with upper and lower case respectively.
\begin{align*} \Delta &= \sqrt{abc(a+b+c)} \\ \text{radius:} \quad r_{\pm} &= \frac{abc}{2\Delta \mp (bc+ca+ab)} \\ \text{centre:} \quad S_{\pm} &= \frac {\left( \frac{\Delta}{a} \mp b \mp c \right) A+ \left( \frac{\Delta}{b} \mp c \mp a \right) B+ \left( \frac{\Delta}{c} \mp a \mp b \right) C } {\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right) \Delta \mp 2(a+b+c)} \end{align*}
You can get the necessary information by substituting $a=1$, $b=2$, $c=3$, $A=(-1,0)$, $B=(2,0)$ and $C=(-1,4)$ as shown in the figure.