Question

What are the coordinates of the circumcenter of a triangle with vertices A(0,1), B(2, 1) , and C(2, 5) ?

Answer 1

Answer: The coordinates of circumcenter is (1,3).

Explanation:

It is given that the triangle have vertices A(0,1), B(2, 1) , and C(2, 5).

The distance formula,

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AB=\sqrt{(2-0)^2+(1-1)^2}=2$

$BC=\sqrt{(2-2)^2+(5-1)^2}=4$

$AC=\sqrt{(2-0)^2+(5-1)^2}=\sqrt{20}$

Since,

$(AC)^2=(AB)^2+(BC)^2$

By pythagoras we can say that the given triangle is a right angle triangle and AC is the hypotenuse of the triangle.

The circumcentre of a right angle triangle is the midpoint of the hypotenuse.

Midpoint of AC,

$\text{Midpoint of AC}=(\frac{0+2}{2}, \frac{1+5}{2})$

$\text{Midpoint of AC}=(\frac{2}{2}, \frac{6}{2})$

$\text{Midpoint of AC}=(1,3)$

Therefore, the coordinates of circumcenter is (1,3).

Answer 2

The correct answer is 
(1, 3) 
you can verify it with
Circumcenter Calculator

http://calculator.tutorvista.com/circumcenter-calculator.html