Find the coordinates of the circum centre of a triangle whose vertices are A(4,6) B(0,4) C(6,2). also finds its circum radius ?

Question

Find the coordinates of the circum centre of a triangle whose
vertices are A(4,6) B(0,4) C(6,2) also finds its circum radius
?

Priyanka Kedia · Accepted Answer

Let O(x,y) be the co-ordinate of the centre of the circumcentre.

Therefore,

$O A = O B \sqrt{{(x - 4)}^{2} + {(y - 6)}^{2}} = \sqrt{{(x - 0)}^{2} + {(y - 4)}^{2}} {(x - 4)}^{2} + {(y - 6)}^{2} = {(x - 0)}^{2} + {(y - 4)}^{2} x^{2} + 16 - 8 x + y^{2} + 36 - 12 y = x^{2} + y^{2} + 16 - 8 y 16 - 8 x + 36 - 12 y = 16 - 8 y - 8 x - 4 y = - 36 2 x + y = 9 ...... (1)$

And,

$O C = O B \sqrt{{(x - 6)}^{2} + {(y - 2)}^{2}} = \sqrt{{(x - 0)}^{2} + {(y - 4)}^{2}} {(x - 6)}^{2} + {(y - 2)}^{2} = {(x - 0)}^{2} + {(y - 4)}^{2} x^{2} + 36 - 12 x + y^{2} + 4 - 4 y = x^{2} + y^{2} + 16 - 8 y 36 - 12 x + 4 - 4 y = 16 - 8 y - 12 x + 4 y = - 24 3 x - y = 6 ...... (2)$

Adding equation (1) and (2), we get,

$5 x = 15 \Rightarrow x = 3$

Substitute value of x in equation (1), we get,

$2 \times 3 + y = 9 \Rightarrow y = 3$

Therefore, co-ordinate of circumcentre is (3,3)

And, radius of the circle OA = $\sqrt{{(3 - 4)}^{2} + {(3 - 6)}^{2}} = \sqrt{1 + 9} = \sqrt{10}$

Therefore, radius of the circle is $\sqrt{10}$ .