Find the coordinates of the circumcentre of the triangle whose vertices are (8, 6), (8, -2) and (2, -2). Also, find its circumradius.

Let O(x, y) be the circumcentre of the circle.

Let A(8, 6), B(8, -2) and C(2, -2) be the vertices of the triangle.

OB = OC [Radii of same circle]

By distance formula,

$\Rightarrow \sqrt{(x - 8)^2 + [y - (-2)]^2} = \sqrt{(x - 2)^2 + [y - (-2)]^2} \\[1em] \Rightarrow \sqrt{x^2 + 64 - 16x + [y + 2]^2} = \sqrt{x^2 + 4 - 4x + [y + 2]^2}$

Squaring both sides we get,

$\Rightarrow x^2 + 64 - 16x + [y + 2]^2 = x^2 + 4 - 4x + [y + 2]^2 \\[1em] \Rightarrow x^2 - x^2 + [y + 2]^2 - [y + 2]^2 + 64 - 4 = 16x - 4x \\[1em] \Rightarrow 12x = 60 \\[1em] \Rightarrow x = \dfrac{60}{12} \\[1em] \Rightarrow x = 5.$

Also,

OA = OB [Radii of same circle]

By distance formula,

$\Rightarrow \sqrt{(x - 8)^2 + (y - 6)^2} = \sqrt{(x - 8)^2 + [y - (-2)]^2} \\[1em] \Rightarrow \sqrt{(x - 8)^2 + y^2 + 36 - 12y} = \sqrt{(x - 8)^2 + [y + 2]^2}$

On squaring both sides,

$\Rightarrow (x - 8)^2 + y^2 + 36 - 12y = (x - 8)^2 + [y + 2]^2 \\[1em] \Rightarrow (x - 8)^2 + y^2 + 36 - 12y = (x - 8)^2 + y^2 + 4 + 4y \\[1em] \Rightarrow (x - 8)^2 - (x - 8)^2 + y^2 - y^2 + 4y + 12y = 36 - 4 \\[1em] \Rightarrow 16y = 32 \\[1em] \Rightarrow y = \dfrac{32}{16} \\[1em] \Rightarrow y = 2.$

O = (x, y) = (5, 2).

Radius = OA.

$OA = \sqrt{(8 - 5)^2 + (6 - 2)^2} \\[1em] = \sqrt{3^2 + 4^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.$

Hence, circumcenter = (5, 2) and circumradius = 5 units.