Circle touching three tangential circles The circles $C_1,C_2$ and $C_3$ with radii $1,2$ and $3$, 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 $C_1,C_2$ and $C_3$ and touches all of them.
I had been trying to do the sum by assuming the centre of the first circle at $(0,1)$ but it did not help.

here is my nice answer: 
let $C$ be the center of enclosed circle then its radius $r$ is given by set formula by HC Rajpoot 
$$r=\frac{abc}{2\sqrt{abc(a+b+c)}+ab+bc+ca}$$
where, $a=1, b=2, c=3$ are radii of three externally touching circles then 
$$r=\frac{1\times2\times 3}{2\sqrt{1\times 2\times 3(1+2+3)}+1\times 2+2\times 3+3\times 1}=\frac{6}{23}$$
now, drop a perpendicular $CN$ of length $y$ from center $C$ to the x-axis to get a right triangle $C_1NC$ in which hypotenuse 
$C_1C=1+\frac6{23}=\frac{29}{23}$. apply pythagorean 
$$C_1N=\sqrt{(C_1C)^2-(CN)^2}=\sqrt{\left(\frac{29}{23}\right)^2-y^2}\ \ \ \ ..........(1)$$
similarly, in right triangle $CNC_2$ in which hypotenuse
$C_2C=2+\frac6{23}=\frac{52}{23}$. apply pythagorean 
$$NC_2=\sqrt{(C_2C)^2-(CN)^2}=\sqrt{\left(\frac{52}{23}\right)^2-y^2}\ \ \ \ ........(2)$$
since, circle $C_1$ & $C_2$ are touching hence, $$C_1N+NC_2=C_1C_2=1+2=3$$
$$\sqrt{\left(\frac{29}{23}\right)^2-y^2}+\sqrt{\left(\frac{52}{23}\right)^2-y^2}=3$$
$$\sqrt{\frac{841}{529}-y^2}=3-\sqrt{\frac{2704}{529}-y^2}$$ taking squares on both sides, i get
$$\sqrt{\frac{2704}{529}-y^2}=\frac{1104}{529}$$
again i take square, 
$$y^2=\frac{2704}{529}-\left(\frac{1104}{529}\right)^2=\frac{400}{529}$$
$$y=\frac{20}{23}$$
above is the correct value of ordinate of center $C$ of the enclosed circle
