Find the value of $k$ so that the pair of straight lines joining the origin to the points of intersection of $\;$ $x^2 + y^2 + 4x - 6y - 36 = 0$ $\;$ and the line $\;$ $y - x - k = 0$ $\;$ may be at right angles.
Equation of given lines is: $\;$ $x^2 + y^2 + 4x - 6y - 36 = 0$ $\;\;\; \cdots \; (1)$
and $\;$ $y - x - k = 0$ $\;\;\; \cdots \; (2)$
The combined equation to the pair of lines joining the origin to the points of intersection of equations $(1)$ and $(2)$ can be obtained by making equation $(1)$ homogeneous in $x$ and $y$ by using equation $(2)$.
Equation $(2)$ can be written as
$y - x = k$
i.e. $\;$ $\dfrac{y - x}{k} = 1$ $\;\;\; \cdots \; (2a)$
$\therefore \;$ Making equation $(1)$ homogeneous in $x$ and $y$ using equation $(2a)$ we get,
$x^2 + y^2 + 4 \left(\dfrac{y - x}{k}\right)x - 6 \left(\dfrac{y - x}{k}\right)y - 36 \times \left(\dfrac{y - x}{k}\right)^2 = 0$
i.e. $\;$ $k^2 x^2 + k^2 y^2 + 4kxy - 4kx^2 -6ky^2 + 6kxy - 36y^2 - 36x^2 + 72xy = 0$
i.e. $\;$ $\left(k^2 -4k - 36\right)x^2 + \left(10k + 72\right)xy + \left(k^2 -6k - 36\right)y^2 = 0$ $\;\;\; \cdots \; (3)$
Equation $(3)$ is a combination of equations $(1)$ and $(2a)$.
$\therefore \;$ Equation $(3)$ will be satisfied by the coordinates of all the points by which equations $(1)$ and $(2a)$ are satisfied simultaneously.
Since equation $(3)$ is homogeneous and of the second degree in $x$ and $y$, it represents a pair of straight lines passing through the origin.
Lines represented by equation $(3)$ will be at right angles if
$\text{coefficient of } x^2 + \text{ coefficient of } y^2 = 0$
i.e. $\;$ $k^2 - 4k - 36 + k^2 - 6k - 36 = 0$
i.e. $\;$ $2k^2 - 10k - 72 = 0$
i.e. $\;$ $k^2 - 5k - 36 = 0$
$\implies$ $k = 9$ $\;$ or $\;$ $k = -4$