Question

Find the coordinate of the circumcentre of the triangle whose vertices are (8,6),(8,-2) and (2,-2). Also, find its circumradius.

• A. Circumradius = 5 units
• B. circumcentre = (2, 5)
• C. Circumradius = 6 units
• D. circumcentre = (5,2)

Answer 1

Answer: (A), (D)

The correct options are
A

Circumradius = 5 units

D

circumcentre = (5,2)

Let A(8,6), B(8,-2) and C(2,-2).

P(x,y) is circumcentre of the circle, then

PA =PB =PC

$P{A}^2=P{B}^2=P{C}^2$

$P{A}^2=P{B}^2$

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

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

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

$\Rightarrow 36-4=16y$

$\Rightarrow 16y=32\Rightarrow y=2.$

P(x,2),C(2,-2),B(8,-2), then

$P{C}^2=(x-2{)}^2+(2+2{)}^2$

$P{B}^2=(x-8{)}^2+(2+2{)}^2$

Since $P{B}^2=P{C}^2$

$\therefore (x-8{)}^2+4^2=(x-2{)}^2+4^2$

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

$\Rightarrow 12x=60\Rightarrow x=5$

$\therefore$ Coordinates of circumcentre are P (5,2).

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

$=\sqrt{9+16}=\sqrt{25}$

=5 units.