Question Number 137374 by oooooooo last updated on 02/Apr/21
Answered by mr W last updated on 02/Apr/21
$${let}\:{OA}={OD}={R}={radius} \\ $$$$\left(\frac{{R}}{\:\sqrt{\mathrm{2}}}−\mathrm{4}\right)^{\mathrm{2}} +\left(\frac{{R}}{\:\sqrt{\mathrm{2}}}+\mathrm{3}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$$\left(\frac{{R}}{\:\sqrt{\mathrm{2}}}−{x}\right)^{\mathrm{2}} +\left(\frac{{R}}{\:\sqrt{\mathrm{2}}}+\mathrm{5}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\left({i}\right): \\ $$$$−\sqrt{\mathrm{2}}{R}+\mathrm{25}=\mathrm{0}\:\Rightarrow{R}=\frac{\mathrm{25}}{\:\sqrt{\mathrm{2}}} \\ $$$$\left({ii}\right): \\ $$$$\left(\frac{\mathrm{25}}{\mathrm{2}}−{x}\right)^{\mathrm{2}} +\left(\frac{\mathrm{25}}{\mathrm{2}}+\mathrm{5}\right)^{\mathrm{2}} =\frac{\mathrm{25}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{25}}{\mathrm{2}}−{x}\right)^{\mathrm{2}} =\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{25}}{\mathrm{2}}−{x}=\pm\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${x}=\frac{\mathrm{25}}{\mathrm{2}}\pm\frac{\mathrm{5}}{\mathrm{2}}=\mathrm{15}\:{or}\:\mathrm{10} \\ $$