Question

The circumcentre of the triangle with vertices $(8,6),(8,-2)$ and $(2,-2)$ is at the point

• A. (2, -1)
• B. (1, -2)
• C. (5, 2)
• D. (2, 5)
• E. (4, 5)

Answer 1

Answer: (C)

Let the vertices of a triangle are $A(8,6),B(8,-2)$ and $C(2,-2)$

Now, $AB=\sqrt{(8-8{)}^2+(6+2{)}^2}$
$=\sqrt{0+8^2}=8$
$BC=\sqrt{(2-8{)}^2+(-2+2{)}^2}=\sqrt{36+0}=6$
and $CA=\sqrt{(8-2{)}^2+(6+2{)}^2}$
$=\sqrt{6^2+8^2}$
$=\sqrt{36+64}=\sqrt{100}=10$
Now, $A{B}^2+B{C}^2=(8{)}^2+(6{)}^2$
$=64+36=100=A{C}^2$
So, $ABC$ is a right angled triangle and right angled at $B$.
We know that, in a right angled triangle, the circumcentre is the mid-point of hypotenuse.
$\therefore$ Mid-point of $AC=\left(\frac{8+2}{2},\frac{6-2}{2}\right)=(5,2)$
Hence, the required circumcentre is $(5,2)$.