Mathematics - Properties of Triangle Question with Solution | TestHub

Circumcentre $P(2,4)$ and orthocenter $O(1,2):O{P}^2=R^2(1-8\cos A\cos B\cos C)$. $PE=R\cos A=\frac{|2(2)-4-3|}{\sqrt{5}}=\frac{3}{\sqrt{5}}$ (distance of P from BC), and $OD=2R\cos B\cos C=\frac{|2(1)-2-3|}{\sqrt{5}}=\frac{3}{\sqrt{5}}$ (distance of O from BC). $\Rightarrow 5=R^2-4\times \frac{3}{\sqrt{5}}\times \frac{3}{\sqrt{5}}\Rightarrow R^2=5+\frac{36}{5}=\frac{61}{5}\Rightarrow R=\sqrt{\frac{61}{5}}$.