Question

Find the co-ordinate of te circumcentre of triangle whose vertices are (8,6),(8-2) and (2,-2). Also its circum radius

Answer 1

$\begin{array}{c}\left(8,6\right),\left(x,y\right)=\left(8,-2\right),\left(x,y\right)\\ \sqrt{{\left(x-8\right)}^2+{\left(y-6\right)}^2}=\sqrt{{\left(x-8\right)}^2+{\left(y+2\right)}^2}\\ {\left(y-6\right)}^2={\left(y+2\right)}^2\\ 36-12y=4y+4\\ 16y=32\\ y=2\\ now\\ \left(2,-2\right),\left(x,y\right)=\left(8,-2\right),\left(x,y\right)\\ \sqrt{{\left(x-2\right)}^2+{\left(y+2\right)}^2}=\sqrt{{\left(x-8\right)}^2+{\left(y+2\right)}^2}\\ {\left(x-2\right)}^2={\left(x-8\right)}^2\\ 12x=60\\ x=5\end{array}$

co-ordinate of circumcentre of triangle $(5,2)$

$\therefore$the circumradius is given by-

$\begin{array}{c}\left(5,2\right)\left(8,6\right)\\ =\sqrt{{\left(5-8\right)}^2+{\left(2-6\right)}^2}\\ =\sqrt{25}\\ =5\end{array}$

Ans $\therefore$ Circumradius=$5$