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We have three circles tangent to each other with radii $1$, $2$, and $3$. Another circle is tangent to the other circles; find the radius of that circle using elementary geometry, without the Soddy Circles formula (aka, Descartes' Theorem).

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My Attempt: I tried to draw a triangle with points $A$, $B$, and $C$, and then I proved that one of its angles is $90^\circ$.

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  • $\begingroup$ I can proof that it is right angle but I want to find radii of small circle $\endgroup$ – Taha Akbari May 28 '16 at 9:18
  • $\begingroup$ Your proof is false: immagine the three outer circles had the same radius, $ABC$ becomes an equilateral triangle. $\endgroup$ – N74 May 28 '16 at 9:38
  • $\begingroup$ OP is right because the radius are given as $1,2,3$ which makes $3,4,5$ triangle. $\endgroup$ – newzad May 28 '16 at 9:47
  • $\begingroup$ Yes that's what I exactly want to say. $\endgroup$ – Taha Akbari May 28 '16 at 9:51
  • $\begingroup$ Do you want a solution using coordinate geometry but without quoting Descartes' Theorem? $\endgroup$ – David Quinn May 28 '16 at 11:11
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As you have observed, $AB$ and $AC$ are perpendicular, so take $A$ as the origin and assign coordinates $B(4,0)$ and $C(0,3)$.

Let the centre of the inner circle be at $(a,b)$ and the radius is $r$. Then, considering distances, the following equations apply:

$$a^2+b^2=(1+r)^2$$ $$(a-4)^2+b^2=(3+r)^2$$ $$a^2+(b-3)^2=(2+r)^2$$

Subtracting the first equation from the second gives rise to $$a=\frac{2-r}{2}$$ Subtracting the first equation from the third gives rise to $$b=\frac{3-r}{3}$$

Substituting these into the first equation then leads to the quadratic equation $$23r^2+132r-36=0$$

The roots are $$r=\frac{6}{23}$$ which is the answer you are looking for, and $$-6$$ which corresponds to the radius $6$ of the outer kissing circle. All of this can be verified by applying Descartes' Formula.

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