# Radius of circle that touches 3 circles, which in turn touch each other

I had $$3$$ circles of radii $$1$$, $$2$$, $$3$$, all touching each other. A smaller circle was constructed such that it touched all the $$3$$ circles.

What is the radius of the smaller circle?

This is what I did:

I conveniently positioned the $$3$$ circles on the coordinate axes and found the coordinates of the centers of each circle.

Then I wrote a general equation of a circle( for the smaller one) , and using the fact that distance between the centers is equal of the sum of radii ( for circles touching each other) I found $$3$$ equations, which I could use to solve for the variables in the general equation. Hence I found the equation of the smaller circle, and thus its radius.

However, I believe that this method is very inefficient as I ended up with so many steps and sub steps to solve the equations.

Is there a better way to approach this?

I would prefer a geometrical solution instead of a coordinate solution.

Thanks for the help!!

Note :

However is there a more efficient solution for this than what was mentioned in those answers?

• I would advise you to have a look at the complex version of Descarte"s theorem here – Jean Marie Jun 12 at 8:42
• But if you want coordinate free proofs,, look at the more general "Apollonius problem" – Jean Marie Jun 12 at 8:47

Let $$A$$, $$B$$ and $$C$$ be centers of circles with radius $$3$$, $$2$$ and $$1$$ respectively and let $$x$$ be a radius of the needed circle with a center $$D$$.

Thus, $$\measuredangle ACB=90^{\circ},$$ $$\cos\measuredangle ACD=\frac{4^2+(1+x)^2-(3+x)^2}{2\cdot4(1+x)}=\frac{2-x}{2(1+x)},$$

$$\cos\measuredangle BCD=\frac{3^2+(1+x)^2-(2+x)^2}{2\cdot3(1+x)}=\frac{3-x}{3(1+x)},$$ which gives $$\left(\frac{2-x}{2(1+x)}\right)^2+\left(\frac{3-x}{3(1+x)}\right)^2=1$$ or $$23x^2+132x-36=0,$$ which gives $$x=\frac{6}{23}.$$

• Thank you so much! – Vamsi Krishna Jun 12 at 8:24

Notice, the radius $$r$$ of the small (inscribed) circle externally touching any three externally kissing (touching) circles of radii $$a, b$$ & $$c$$ is given by the generalized formula as follows

$$\boxed{\color{blue}{r=\frac{abc}{2\sqrt{abc(a+b+c)}+ab+bc+ca}}}$$ Now, substituting the values of radii of three externally touching circle i.e. $$a=1, b=2$$ & $$c=3$$ in above generalized formula, we get radius of small circle

$$r=\frac{1\cdot 2\cdot 3}{2\sqrt{1\cdot2\cdot 3(1+2+3)}+1\cdot 2+2\cdot 3+3\cdot 1}$$ $$r=\color{blue}{\frac{6}{23}}$$

• Thanks a lot!!! – Vamsi Krishna Jun 12 at 8:24