Find the center and radius of the circle whose equation is $\;$ $\left(x - 1\right) \left(x - 3\right) + \left(y - 2\right) \left(y - 4\right) = 0$
Given equation of circle: $\;$ $\left(x - 1\right) \left(x - 3\right) + \left(y - 2\right) \left(y - 4\right) = 0$
i.e. $\;$ $x^2 - 4x + 3 + y^2 - 6y + 8 = 0$
i.e. $\;$ $x^2 - 4x + y^2 - 6y + 11 = 0$
Completing the squares, we get
$\left(x^2 - 4x + 4\right) + \left(y^2 - 6y + 9\right) + 11 - 4 - 9 = 0$
i.e. $\;$ $\left(x - 2\right)^2 + \left(y - 3\right)^2 = 2$
Comparing with the standard equation of circle $\;$ $\left(x - h\right)^2 + \left(y - k\right)^2 = r^2$, $\;$ we have
$h = 2, \; \; k = 3, \;\; r^2 = 2$
center of the circle $= \left(h, k\right) = \left(2, 3\right)$
radius of the circle $= r = \sqrt{2}$