Find the coordinates of the circumcentre of the triangle whose vertices are $(8, 6)$ , $(8, - 2)$ and $(2, - 2)$ also, find its circum-radius.

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Solution

Let the circumcentre be $(x, y)$
$A (8, 6), B (8, - 2), C (2, - 2)$
$S A = S B = S C = R$
So $S A = S B$
$\sqrt{(8 - x)^{2} + (6 - y)^{2}} = \sqrt{(8 - x)^{2} + (- 2 - y)^{2}}$
$x^{2} + y^{2} + 64 + 36 - 16 x - 12 y = x^{2} + y^{2} - 16 x + 4 y + 64 + 4$
$100 = 16 y + 68$
$y = 2$
Similiarly $S B = S C$
$\sqrt{(8 - x)^{2} + (- 2 - y)^{2}} = \sqrt{(2 - x)^{2} + (- 2 - y)^{2}}$
$x^{2} + y^{2} - 16 x + 4 y + 64 + 4 = x^{2} + y^{2} - 4 y - 4 x + 8$
$- 12 x = - 60 \Rightarrow x = 5$
So circumcenter SA= $\sqrt{(8 - 5)^{2} + (6 - 2)^{2}} = 5$

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