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, $A B=\sqrt{(8-8)^{2}+(6+2)^{2}}$
$=\sqrt{0+8^{2}}=8$
$B C=\sqrt{(2-8)^{2}+(-2+2)^{2}}=\sqrt{36+0}=6$
and $C A=\sqrt{(8-2)^{2}+(6+2)^{2}}$
$=\sqrt{6^{2}+8^{2}}$
$=\sqrt{36+64}=\sqrt{100}=10$
Now, $ A B^{2}+B C^{2} =(8)^{2}+(6)^{2} $
$ =64+36=100=A C^{2}$
So, $A B C$ 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 $A C=\left(\frac{8+2}{2}, \frac{6-2}{2}\right)=(5,2)$
Hence, the required circumcentre is $(5,2)$.