Question

Find the coordinates of the circumcentre of the triangle whose vertices are (8, 6), (8, – 2) and (2, – 2). Also, find its circumradius. [3 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark
Calculation : 1 Mark

Recall that the circumcentre of a traingle is equidistatnt 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 ,

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

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

$\Rightarrow x^2+y^2-16x-12y+100=x^2+y^2-16x+4y+68$

$\Rightarrow 16y=32$

$\Rightarrow y=2$

and

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

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

$\Rightarrow x^2+y^2-16x+4y+68=x^2+y^2-4x+4y+8$

$\Rightarrow 12x=60$

$\Rightarrow x=5$

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

Also, Circum- radius = PA = PB = PC

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

$=\sqrt{9+16}$

$=5$