Using distance formula, show that (3, 3) is the center of the circle passing through the points (6, 2), (0, 4) and (4, 6).

Given,

Center (O) = (3, 3)

Let, A = (6, 2), B = (0, 4) and C = (4, 6).

By distance formula,

$d = \sqrt{(x$2 - x1)^2 + (y2 - y1)^2} \\[1em] \Rightarrow AO = \sqrt{(6 - 3)^2 + (2 - 3)^2} \\[1em] = \sqrt{(3)^2 + (-1)^2} \\[1em] = \sqrt{9 + 1} \\[1em] = \sqrt{10}. \\[1em] \Rightarrow BO = \sqrt{(0 - 3)^2 + (4 - 3)^2} \\[1em] = \sqrt{(-3)^2 + 1^2} \\[1em] = \sqrt{9 + 1} \\[1em] = \sqrt{10}. \\[1em] \Rightarrow CO = \sqrt{(4 - 3)^2 + (6 - 3)^2} \\[1em] = \sqrt{1^2 + 3^2} \\[1em] = \sqrt{1 + 9} \\[1em] = \sqrt{10}. \\[1em]

Since, AO = BO = CO.

Hence, proved that (3, 3) is the center of the circle passing through the points (6, 2), (0, 4) and (4, 6).