Find the equation of the circle which passes through the points of intersection of $\;$ $x^2 + y^2 - 4x - 6y - 21 = 0$ $\;$ and $\;$ $3x + 4y + 5 = 0$ $\;$ and the point $\;$ $\left(1, 2\right)$.
Given circle: $\;\;$ $x^2 + y^2 - 4x - 6y - 21 = 0$ $\;\;\; \cdots \; (1)$
Given line: $\;\;$ $3x + 4y + 5 = 0$ $\;\;\; \cdots \; (2)$
Equation of circle passing through the intersection of equations $(1)$ and $(2)$ is
$\left(x^2 + y^2 - 4x - 6y - 21\right) + \lambda \left(3x + 4y + 5\right) = 0$ $\;\;$ where $\;$ $\lambda$ $\;$ is a constant.
i.e. $\;$ $x^2 + y^2 + \left(3 \lambda - 4\right) x + \left(4 \lambda - 6\right) y + 5 \lambda - 21 = 0$ $\;\;\; \cdots \; (3)$
Given: Equation $(3)$ passes through the point $\;$ $\left(1, 2\right)$
$\therefore \;$ We have
$1^2 + 2^2 + \left(3 \lambda - 4\right) \times 1 + \left(4 \lambda - 6\right) \times 2 + 5 \lambda - 21 = 0$
i.e. $\;$ $5 + 3 \lambda - 4 + 8 \lambda - 12 + 5 \lambda - 21 = 0$
i.e. $\;$ $16 \lambda - 32 = 0$ $\implies$ $\lambda = 2$ $\;\;\; \cdots \; (4)$
In view of equation $(4)$, equation $(3)$ becomes
$x^2 + y^2 + 2x + 2y - 11 = 0$
which is the required equation of circle.