Find the equation to the circle whose diameter is the intercept made on $\;$ $3x + 4y = 1$ $\;$ by $\;$ $5x^2 + 6xy + y^2 = 0$.
Equation of given line: $\;$ $3x + 4y = 1$ $\;\;\; \cdots \; (1)$
Equation of given pair of lines: $\;$ $5x^2 + 6 xy + y^2 = 0$ $\;\;\; \cdots \; (2)$
i.e. $\;$ $5x^2 + 5xy + xy + y^2 = 0$
i.e. $\;$ $5x \left(x + y\right) + y \left(x + y\right) = 0$
i.e. $\;$ $\left(5x + y\right) \left(x + y\right) = 0$
i.e. $\;$ Equation $(2)$ represents the lines
$5x + y = 0$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $x + y = 0$ $\;\;\; \cdots \; (2b)$
Solving equations $(1)$ and $(2a)$ simultaneously gives the point $\;$ $P \left(x_1, y_1\right) = \left(\dfrac{-1}{17}, \dfrac{5}{17}\right)$
Solving equations $(1)$ and $(2b)$ simultaneously gives the point $\;$ $Q \left(x_2, y_2\right) = \left(-1, 1\right)$
As per the problem, points $P$ and $Q$ are the extremities of the diameter of the required circle.
$\therefore \;$ Equation of required circle is
$\left(x - x_1\right) \left(x - x_2\right) + \left(y - y_1\right) \left(y - y_2\right) = 0$
i.e. $\;$ $\left(x + \dfrac{1}{17}\right) \left(x + 1\right) + \left(y - \dfrac{5}{17}\right) \left(y - 1\right) = 0$
i.e. $\;$ $x^2 + \dfrac{18}{17} x + \dfrac{1}{17} + y^2 - \dfrac{22}{17} y + \dfrac{5}{17} = 0$
i.e. $\;$ $17 x^2 + 17 y^2 + 18 x - 22 y + 6 = 0$