Question

Find the coordinates of the circumcenter of a triangle whose vertices are (8, 6), (8, – 2) and (2, – 2). What is the circumradius of this triangle?

Answer 1

Recall that the circumcentre of a traingle is equidistant from the verices of a traingle.
Let A (8 , 6 ), B ( 8, -2 ) and C (2 , -2) be the vertices of the given traingle and let P ( x, y) be the circumcentre of this traingle .
Then, PA = PB = PC
$\Rightarrow$ ${PA}^2$ = ${PB}^2$ = ${PC}^2$

Now , ${PA}^2$ = ${PB}^2$

$\Rightarrow$ ${(x-8)}^2$ + ${(y-6)}^2$ = ${(x-8)}^2$ + ${(y+2)}^2$

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

$\Rightarrow$ 16y = 32

y = 2

and

${PB}^2$ = ${PC}^2$

$\Rightarrow$ ${(x-8)}^2$ + ${(y+2)}^2$ = ${(x-2)}^2$ + ${(y+2)}^2$

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

$\Rightarrow$ 12x = 60

$\Rightarrow$ x = 5

So the coordinates of the circumcentre P are (5 ,2)

Also, Circumradius = PA = PB = PC

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

= $\sqrt{9+16}$

= 5 units