Descartes' Circle Theorem

$\displaystyle \alpha +\beta +\gamma =2\pi$

$\displaystyle \cos 2\alpha +\cos 2\beta +\cos 2\gamma$

$\displaystyle =2{{\cos }^{2}}\alpha -1+2{{\cos }^{2}}\beta -1+2{{\cos }^{2}}\gamma -1$

$\displaystyle =2({{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma )-3\ \ \ .............(1)$

$\displaystyle \cos 2\alpha +\cos 2\beta +\cos 2\gamma$

$\displaystyle =2\cos (\alpha +\beta )\cos (\alpha -\beta )+2{{\cos }^{2}}\gamma -1$

$\displaystyle =2\cos (2\pi -\gamma )\cos (\alpha -\beta )+2{{\cos }^{2}}\gamma -1$

$\displaystyle =2\cos \gamma \cos (\alpha -\beta )+2{{\cos }^{2}}\gamma -1$

$\displaystyle =2\cos \gamma \ [\ \cos (\alpha -\beta )+\cos \gamma \ ]-1\ \ $

$\displaystyle =2\cos \gamma \ [\ \cos (\alpha -\beta )+\cos (\alpha +\beta )\ ]-1\ $

$\displaystyle =4\cos \alpha \cos \beta \cos \gamma -1\ \ \ \ \ .........................(2)$

$\displaystyle \text{From}\ (1)\ \text{and}\ \ (2)\ ,$

$\displaystyle 2({{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma )-3=4\cos \alpha \cos \beta \cos \gamma -1\ $

$\displaystyle \therefore \ {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1+2\cos \alpha \cos \beta \cos \gamma$

$\displaystyle Let\ \angle AOB=\gamma \ ,\ \angle BOC=\alpha \ \ ,\ \angle AOC=\beta$

$\displaystyle \text{By}\ \text{the}\ \text{law}\ \text{of}\ \text{cosines}\ ,$

$\displaystyle \cos \gamma =\frac{{O{{A}^{2}}+O{{B}^{2}}-A{{B}^{2}}}}{{2\ OA\ .\ OB}}$

$\displaystyle \ \ \ \ \ \ \ =\frac{{{{{({{r}_{4}}+{{r}_{1}})}}^{2}}+{{{({{r}_{4}}+{{r}_{2}})}}^{2}}-{{{({{r}_{1}}+{{r}_{2}})}}^{2}}}}{{2\ ({{r}_{4}}+{{r}_{1}})\ ({{r}_{4}}+{{r}_{2}})}}$

$\displaystyle \ \ \ \ \ \ \ =\frac{{2{{r}_{4}}^{2}+2{{r}_{4}}({{r}_{1}}+{{r}_{2}})-2{{r}_{1}}{{r}_{2}}}}{{2\ ({{r}_{4}}+{{r}_{1}})\ ({{r}_{4}}+{{r}_{2}})}}$

$\displaystyle \ \ \ \ \ \ \ =1-\frac{{2{{r}_{1}}{{r}_{2}}}}{{({{r}_{4}}+{{r}_{1}})\ ({{r}_{4}}+{{r}_{2}})}}$

$\displaystyle \cos \gamma =1-\frac{{2{{k}_{4}}^{2}}}{{({{k}_{4}}+{{k}_{1}})({{k}_{4}}+{{k}_{2}})}}=1-{{\lambda }_{3}}$

$\displaystyle \text{Similarly}\ \text{,}\ \text{we}\ \text{have}$

$\displaystyle \cos \beta =1-\frac{{2{{k}_{4}}^{2}}}{{({{k}_{4}}+{{k}_{1}})({{k}_{4}}+{{k}_{3}})}}=1-{{\lambda }_{2}}$

$\displaystyle \cos \alpha =1-\frac{{2{{k}_{4}}^{2}}}{{({{k}_{4}}+{{k}_{2}})({{k}_{4}}+{{k}_{3}})}}=1-{{\lambda }_{1}}$

$\displaystyle {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1+2\cos \alpha \cos \beta \cos \gamma$

$\displaystyle {{(1-{{\lambda }_{1}})}^{2}}+{{(1-{{\lambda }_{2}})}^{2}}+{{(1-{{\lambda }_{3}})}^{2}}=1+2(1-{{\lambda }_{1}})(1-{{\lambda }_{2}})(1-{{\lambda }_{3}})$

$\displaystyle \text{Simplifying}\ \text{,}\ \text{we}\ \text{get}$

$\displaystyle {{\lambda }_{1}}^{2}+{{\lambda }_{2}}^{2}+{{\lambda }_{3}}^{2}+2{{\lambda }_{1}}{{\lambda }_{2}}{{\lambda }_{3}}=2({{\lambda }_{1}}{{\lambda }_{2}}+{{\lambda }_{2}}{{\lambda }_{3}}+{{\lambda }_{1}}{{\lambda }_{3}})$

$\displaystyle \text{Dividing}\ \text{both}\ \text{sides}\ by\ {{\lambda }_{1}}{{\lambda }_{2}}{{\lambda }_{3}}\ \text{gives}\ $

$\displaystyle \frac{{{{\lambda }_{1}}}}{{{{\lambda }_{2}}{{\lambda }_{3}}}}+\frac{{{{\lambda }_{2}}}}{{{{\lambda }_{1}}{{\lambda }_{3}}}}+\frac{{{{\lambda }_{3}}}}{{{{\lambda }_{1}}{{\lambda }_{2}}}}+2=2(\frac{1}{{{{\lambda }_{1}}}}+\frac{1}{{{{\lambda }_{2}}}}+\frac{1}{{{{\lambda }_{3}}}})$

$\displaystyle \text{Putting}\ \text{the}\ \text{values}\ \text{of}\ {{\lambda }_{1}}\ ,\ {{\lambda }_{2}}\ ,\ {{\lambda }_{3}}\ \ \text{gives}$

$\displaystyle \frac{{{{{({{k}_{4}}+{{k}_{1}})}}^{2}}}}{{2{{k}_{4}}^{2}}}+\frac{{{{{({{k}_{4}}+{{k}_{2}})}}^{2}}}}{{2{{k}_{4}}^{2}}}+\frac{{{{{({{k}_{4}}+{{k}_{3}})}}^{2}}}}{{2{{k}_{4}}^{2}}}+2$

$\displaystyle =2\ [\ \frac{{({{k}_{4}}+{{k}_{2}})({{k}_{4}}+{{k}_{3}})}}{{2{{k}_{4}}^{2}}}+\frac{{({{k}_{4}}+{{k}_{1}})({{k}_{4}}+{{k}_{3}})}}{{2{{k}_{4}}^{2}}}+\frac{{({{k}_{4}}+{{k}_{1}})({{k}_{4}}+{{k}_{2}})}}{{2{{k}_{4}}^{2}}}\ ]$

$\displaystyle {{k}_{1}}^{2}+{{k}_{2}}^{2}+{{k}_{3}}^{2}+2{{k}_{4}}({{k}_{1}}+{{k}_{2}}+{{k}_{3}})+7{{k}_{4}}^{2}=6{{k}_{4}}^{2}+4{{k}_{4}}({{k}_{1}}+{{k}_{2}}+{{k}_{3}})+2({{k}_{1}}{{k}_{2}}+{{k}_{2}}{{k}_{3}}+{{k}_{1}}{{k}_{3}})$

$\displaystyle {{k}_{1}}^{2}+{{k}_{2}}^{2}+{{k}_{3}}^{2}+{{k}_{4}}^{2}=2{{k}_{4}}({{k}_{1}}+{{k}_{2}}+{{k}_{3}})+2({{k}_{1}}{{k}_{2}}+{{k}_{2}}{{k}_{3}}+{{k}_{1}}{{k}_{3}})$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{({{k}_{1}}+{{k}_{2}}+{{k}_{3}}+{{k}_{4}})}^{2}}-({{k}_{1}}^{2}+{{k}_{2}}^{2}+{{k}_{3}}^{2}+{{k}_{4}}^{2})$

$\displaystyle \therefore \ {{({{k}_{1}}+{{k}_{2}}+{{k}_{3}}+{{k}_{4}})}^{2}}=2({{k}_{1}}^{2}+{{k}_{2}}^{2}+{{k}_{3}}^{2}+{{k}_{4}}^{2})$

In the figure below , semicircles with centres at A and B and with radii 2 and 1 respectively,are drawn in the interior of , and sharing bases with , a semicircle with diameter JK . The two smaller semicircles are externally tangent to each other and internally tangent to each other and internally tangent to the largest semicircle . A circle centred at P is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle . What is the radius of the circle centred at P .
 

Solution
$\displaystyle {{r}_{1}}=1\ ,\ {{r}_{2}}\ =2\ ,\ {{r}_{3}}=3$

$\displaystyle {{k}_{1}}=\frac{1}{{{{r}_{1}}}}=1\ ,\ {{k}_{2}}=\frac{1}{{{{r}_{2}}}}=\frac{1}{2}\ \ ,\ {{k}_{3}}=-\frac{1}{{{{r}_{3}}}}=-\frac{1}{3}$

By Descartes' Circle Theorem ,

$\displaystyle {{({{k}_{1}}+{{k}_{2}}+{{k}_{3}}+{{k}_{4}})}^{2}}=2({{k}_{1}}^{2}+{{k}_{2}}^{2}+{{k}_{3}}^{2}+{{k}_{4}}^{2})$

$\displaystyle {{(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{{{{r}_{4}}}})}^{2}}=2(1+\frac{1}{4}+\frac{1}{9}+\frac{1}{{{{r}_{4}}^{2}}})$

$\displaystyle {{(\frac{{7{{r}_{4}}+6}}{{6{{r}_{4}}}})}^{2}}=2(\frac{{49{{r}_{4}}^{2}+36}}{{36{{r}_{4}}^{2}}})$

$\displaystyle 49{{r}_{4}}^{2}+84{{r}_{4}}+36=98{{r}_{4}}^{2}+72$

$\displaystyle 49{{r}_{4}}^{2}-84{{r}_{4}}+36=0$

$\displaystyle {{(7{{r}_{4}}-6)}^{2}}=0$

$\displaystyle \ \ \ \ 7{{r}_{4}}-6=0$

$\displaystyle \ \ \ \ \therefore \ \ \ {{r}_{4}}=\frac{6}{7}$