Question Number 175166 by cherokeesay last updated on 21/Aug/22
Answered by ajfour last updated on 22/Aug/22
$${let}\:{centre}\:{of}\:{required}\:{circle} \\ $$$${be}\:\:\left({h},{R}\right) \\ $$$${h}=−\frac{{r}}{\mathrm{2}\sqrt{\mathrm{2}}}=−\mathrm{4} \\ $$$$\Rightarrow\:\:{r}=\mathrm{8}\sqrt{\mathrm{2}} \\ $$$$\left({h}−\frac{{r}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\left(\frac{{r}}{\:\sqrt{\mathrm{2}}}−{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{16}+\left(\mathrm{8}−{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\mathrm{8}\left(\mathrm{2}{R}−\mathrm{8}\right)=\mathrm{16} \\ $$$${R}=\mathrm{5} \\ $$