Question

Find the circumradius of the triangle whose vertices are (8, 6), (8, – 2) and (2, – 2).

• A. 2.5
• B. 5
• C. 7.5
• D. 10

Answer 1

The correct option is B

5

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

$\Rightarrow$ 16y = 32

y = 2

and

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

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

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