Find the equation of the circle whose diameter is the common chord of the circles $x^2 + y^2 + 2x + 3y + 1 = 0$ and $x^2 + y^2 + 4x + 3y + 2 = 0$
Given circles:
$x^2 + y^2 + 2x + 3y + 1 = 0$ $\;\;\; \cdots \; (1)$
$x^2 + y^2 + 4x + 3y + 2 = 0$ $\;\;\; \cdots \; (2)$
Comparing equation $(1)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0$ $\;$ gives
$g_1 = 1, \; f_1 = \dfrac{3}{2}, \; c_1 = 1$
Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0$ $\;$ gives
$g_2 = 2, \; f_2 = \dfrac{3}{2}, \; c_2 = 2$
Common chord of circles $(1)$ and $(2)$ is
$2x \left(g_1 - g_2\right) + 2y \left(f_1 - f_2\right) + c_1 - c_2 = 0$
i.e. $\;$ $2x \left(1 - 2\right) + 2y \left(\dfrac{3}{2} - \dfrac{3}{2}\right) + 1 - 2 = 0$
i.e. $\;$ $-2x -1 = 0$
i.e. $\;$ $x = \dfrac{-1}{2}$
Substituting the value of $x$ in equation $(1)$ gives
$\dfrac{1}{4} + y^2 - 1 + 3y + 1 = 0$
i.e. $\;$ $y^2 + 3y + \dfrac{1}{4} = 0$
i.e. $\;$ $4y^2 + 12 y + 1 = 0$
Solving the quadratic equation gives $\;$ $y = \dfrac{-3}{2} \pm \sqrt{2}$
$\therefore \;$ As per question, the coordinates of the end points of the diameter of the required circle are
$A \left(x_1, y_1\right) = \left(\dfrac{-1}{2}, \dfrac{-3}{2} + \sqrt{2}\right)$ $\;$ and $\;$ $B \left(x_2, y_2\right) = \left(\dfrac{-1}{2}, \dfrac{-3}{2} - \sqrt{2}\right)$
$\therefore \;$ Equation equation of the 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. $\;$ $\left(x - \dfrac{1}{2}\right) \left(x - \dfrac{1}{2}\right) + \left(y + \dfrac{3}{2} - \sqrt{2}\right) \left(y + \dfrac{3}{2} + \sqrt{2}\right) = 0$
i.e. $\;$ $x^2 - x + \dfrac{1}{4} + y^2 + \dfrac{9}{4} + 3y - 2 = 0$
i.e. $\;$ $x^2 + y^2 - x + 3y + \dfrac{1}{2} = 0$
i.e. $\;$ $2x^2 + 2y^2 - 2x + 6y + 1 = 0$