Find the coordinates of the midpoint of the chord which the circle $x^2 + y^2 + 4x - 2y - 3 = 0$ cuts off on the line $y = x + 2$
Given equation of circle: $\hspace{1em}$ $x^2 + y^2 + 4x - 2y - 3 = 0$ $\;\;\; \cdots \; (1)$
Comparing with the standard equation: $\hspace{1em}$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives
$2g = 4 \implies g = 2$, $\;\;$ $2f = -2 \implies f = -1$, $\;\;$ $c = -3$
$\therefore$ $\;$ Center of circle $= C = \left(-g, -f\right) = \left(-2,1\right)$
Equation of line is: $\hspace{1em}$ $y = x + 2$ $\;\;\; \cdots \; (2)$
Let the given line cut the circle at points $A$ and $B$.
Draw $CM$ perpendicular to chord $AB$.
Then $M$ is the midpoint of $AB$.
Slope of given line $= m = 1$
$\therefore$ $\;$ Slope of $CM = \dfrac{-1}{m}= -1$
$\therefore$ $\;$ Equation of CM is: $\hspace{1em}$ $\left(y - 1\right) = -1 \left(x + 2\right)$
i.e. $-y = x + 1$ $\;\;\; \cdots \; (3)$
Adding equations $(2)$ and $(3)$ gives
$0 = 2x + 3$ $\implies$ $x = \dfrac{-3}{2}$
$\therefore$ $\;$ We have from equation $(2)$, $\hspace{1em}$ $y = \dfrac{-3}{2} + 2 = \dfrac{1}{2}$
$\therefore$ $\;$ Coordinates of point $M$ are $\left(\dfrac{-3}{2}, \dfrac{1}{2}\right)$