Find the equation of the line which is inclined at $45^\circ$ to the positive direction of the $X$ axis and on which the circles $x^2 + y^2 = 9$ and $x^2 + y^2 - 12x - 8y - 29 = 0$ cut equal intercepts.
Given equations of circles:
$x^2 + y^2 = 9$ $\;\;\; \cdots \; (1)$
$x^2 + y^2 - 12x - 8y - 29 = 0$ $\;\;\; \cdots \; (2)$
Let the equation of the required line be: $\;\;$ $y = mx + c$
Since the required line makes an angle of $45^\circ$ with the positive direction of the $X$ axis,
$\implies$ slope of the line $= m = \tan 45^\circ = 1$
$\therefore \;$ Equation of the required line is: $\;\;$ $y = x + c$ $\;\;\; \cdots \; (3)$
Let equations $(1)$ and $(3)$ intersect at the points $A$ and $B$.
We have, from equations $(1)$ and $(3)$
$x^2 + \left(x + c\right)^2 = 9$
i.e. $\;$ $2 x^2 + 2 c x + c^2 - 9 = 0$
$\implies$ $x = \dfrac{-2c \pm \sqrt{4 c^2 - 4 \times 2 \times \left(c^2 - 9\right)}}{2 \times 2}$
i.e. $\;$ $x = \dfrac{-2c \pm \sqrt{4 c^2 - 8c^2 + 72}}{4}$
i.e. $\;$ $x = \dfrac{-2c \pm \sqrt{72 - 4 c^2}}{4}$
i.e. $\;$ $\dfrac{-2 c \pm 2 \sqrt{18 - c^2}}{4}$
i.e. $\;$ $x = \dfrac{-c \pm \sqrt{18 - c^2}}{2}$
Substituting the value of $x$ in equation $(3)$ gives
$y = \dfrac{-c \pm \sqrt{18 - c^2}}{2} + c$
i.e. $y = \dfrac{c \pm \sqrt{18 - c^2}}{2}$
$\therefore \;$ The points of intersections are
$A = \left(\dfrac{-c + \sqrt{18 - c^2}}{2}, \dfrac{c + \sqrt{18 - c^2}}{2}\right)$, $\;$ $B = \left(\dfrac{-c - \sqrt{18 - c^2}}{2}, \dfrac{c - \sqrt{18 - c^2}}{2}\right)$
$AB$ is the length of the chord.
$\therefore \;$ $AB^2 = \left[\dfrac{-c - \sqrt{18 - c^2}}{2} - \left(\dfrac{c + \sqrt{18 - c^2}}{2}\right)\right]^2 + \left[\dfrac{c - \sqrt{18 - c^2}}{2} - \dfrac{c + \sqrt{18 - c^2}}{2}\right]^2$
i.e. $\;$ $AB^2 = 18 - c^2 + 18 - c^2$
i.e. $\;$ $AB^2 = 36 - 2c^2$ $\;\;\; \cdots \; (4)$
Let equations $(2)$ and $(3)$ intersect at the points $C$ and $D$.
Solving equations $(2)$ and $(3)$ simultaneously gives,
$x^2 + \left(x + c\right)^2 - 12 x - 8 \left(x + c\right) - 29 = 0$
i.e. $\;$ $x^2 + x^2 + 2cx + c^2 - 12x - 8x - 8c - 29 = 0$
i.e. $\;$ $2x^2 + \left(2c - 20\right)x + c^2 - 8c - 29 = 0$
i.e. $\;$ $x = \dfrac{20 - 2c \pm \sqrt{\left(2c - 20\right)^2 - 4 \times 2 \times \left(c^2 - 8c - 29\right)}}{2 \times 2}$
i.e. $\;$ $x = \dfrac{20 - 2c \pm \sqrt{4c^2 + 400 - 80c - 8c^2 + 64c + 232}}{4}$
i.e. $\;$ $x = \dfrac{20 - 2c \pm \sqrt{632 - 16c - 4c^2}}{4}$
i.e. $\;$ $x = \dfrac{10 - c \pm \sqrt{158 - 4c - c^2}}{2}$
Substituting the value of $x$ in equation $(3)$ gives,
$y = \dfrac{10 -c \pm \sqrt{158 - 4c - c^2}}{2} + c$
i.e. $y = \dfrac{10 + c \pm \sqrt{158 - 4c - c^2}}{2}$
$\therefore \;$ the points of intersection are
$C = \left(\dfrac{10 - c + \sqrt{158 - 4c - c^2}}{2}, \dfrac{10 + c + \sqrt{158 - 4c - c^2}}{2}\right)$ $\;$ and
$D = \left(\dfrac{10 - c - \sqrt{158 - 4c - c^2}}{2}, \dfrac{10 + c - \sqrt{158 - 4c - c^2}}{2}\right)$
$CD$ is the length of the chord.
$\therefore \;$ $CD^2 = \left(\dfrac{10 - c + \sqrt{158 - 4c - c^2}}{2} - \dfrac{10 - c - \sqrt{158 - 4c - c^2}}{2}\right)^2$
$\hspace{1.5cm}$ $+ \left(\dfrac{10 + c + \sqrt{158 - 4c - c^2}}{2} - \dfrac{10 + c - \sqrt{158 - 4c - c^2}}{2}\right)^2$
i.e. $\;$ $CD^2 = 158 - 4c - c^2 + 158 - 4c - c^2$
i.e. $\;$ $CD^2 = 316 - 8c - 2c^2$ $\;\;\; \cdots \; (5)$
Since the required line cuts equal intercepts on both the circles, we have
$AB = CD$
$\implies$ $AB^2 = CD^2$
i.e. $\;$ $36 - 2 c^2 = 316 - 8c - 2c^2$
i.e. $\;$ $8c = 280$ $\implies$ $c = 35$
$\therefore \;$ The required equation of line is
$y = x + 35$