Question

Find the coordiantes of the circumradius of the triangle whose vertices are $(8,6)$,$(8,2)$ and $(2,2)$.Also, find its circumradius.

Answer 1

Let the coordinates of the circumcentre of the triangle be $(x,y)$.

Circumcentre of a triangle is equidistant from each of the vertices.

Distance between $(8,6)$ and $(x,y)=$ Distance between $(8,-2)$ and $(x,y)$

$[{(x-8)}^2+{(y-6)}^2]=[{(x-8)}^2+{(y+2)}^2]$

${(y-6)}^2={(y+2)}^2$

$y^2+36-12y=y^2+4y+4$

$36-12y=4y+4$

$16y=32$

$y=2$

Distance between $(2,-2)$ and $(x,y)=$ Distance between $(8,-2)$ and $(x,y)$

$\sqrt{[{(x-2)}^2+{(y+2)}^2]}=\sqrt{[{(x-8)}^2+{(y+2)}^2]}$

$[{(x-2)}^2+{(y+2)}^2]=[{(x-8)}^2+{(y+2)}^2]$

${(x-2)}^2={(x-8)}^2$

$x^2+4-4x=x^2-16x+64$

$4-4x=-16x+64$

$12x=60$

$x=5$

Hence, the coordinates of the circumcentre of the triangle are $(5,2)$.

Circumradius $=\sqrt{[{(5-8)}^2+{(2-6)}^2]}$

$=\sqrt{(9+16)}$

$=\sqrt{25}$

$=5$ units.