Find the equation of the circle on the line joining the points $\left(1, 5\right)$, $\;$ $\left(3, 1\right)$ as diameter and find the extremities of the perpendicular diameter.
Given points: $\;$ $A\left(x_1, y_1\right) = \left(1, 5\right)$, $\;$ $B \left(x_2, y_2\right) = \left(3, 1\right)$
Equation of circle in terms of ends of diameter $A \left(x_1, y_1\right)$ and $B \left(x_2, y_2\right)$ is
$\left(x - x_1\right) \left(x - x_2\right) + \left(y - y_1\right) \left(y - y_2\right) = 0$
$\therefore \;$ The required equation of circle is
$\left(x - 1\right) \left(x - 3\right) + \left(y - 5\right) \left(y - 1\right) = 0$
i.e. $\;$ $x^2 - 4x + 3 + y^2 - 6y + 5 = 0$
i.e. $\;$ $x^2 + y^2 - 4x - 6y + 8 = 0$ $\;\;\; \cdots \; (1)$
Comparing equation $(1)$ with standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives
$g = -2, \; f = -3, \; c = 8$
$\therefore \;$ Center of circle given by equation $(1)$ is
$C \left(-g, -f\right) = \left(2, 3\right)$
Slope of diameter $AB = \dfrac{1 - 5}{3 - 1} = -2$
Let $CD$ be the diameter perpendicular to $AB$.
Let equation of $CD$ be $\;$ $y = mx + c$ $\;\;\; \cdots \; (2)$
Then slope of $CD = m = \dfrac{-1}{\text{slope of AB}} = \dfrac{1}{2}$
$\therefore \;$ Equation of $CD$ [equation $(2)$] becomes $\;\;$ $y = \dfrac{1}{2} x + c$ $\;\;\; \cdots \; (3)$
Since $CD$ passes through the point $C \left(2, 3\right)$, equation $(3)$ becomes
$3 = \dfrac{2}{2} + c$ $\implies$ $c = 2$
$\therefore \;$ Equation of $CD$ [equation $(3)$] is
$y = \dfrac{1}{2} x + 2$
i.e. $\;$ $x = 2y - 4$ $\;\;\; \cdots \; (4)$
Solving equations $(1)$ and $(4)$ simultaneously gives
$\left(2y - 4\right)^2 + y^2 - 4 \left(2y - 4\right) - 6y + 8 = 0$
i.e. $\;$ $4y^2 - 16 y + 16 + y^2 - 8y + 16 - 6y + 8 = 0$
i.e. $\;$ $5y^2 - 30y + 40 = 0$
i.e. $\;$ $y^2 - 6y + 8 = 0$
i.e. $\;$ $\left(y - 4\right) \left(y - 2\right) = 0$
i.e. $\;$ $y = 4$ $\;$ or $\;$ $y = 2$
$\therefore \;$ We have from equation $(4)$,
when $\;$ $y = 4$, $\;$ $x = 2 \times 4 - 4 = 4$
when $\;$ $y = 2$, $\;$ $x = 2 \times 2 - 4 = 0$
$\therefore \;$ The extremities of diameter $CD$ are $\left(4, 4\right)$ and $\left(0,2\right)$