Find the equation of the circle with center $\left(2,3\right)$ and passing through the intersection of the lines $\;$ $3x - 2y -1 = 0$ and $4x + y - 27 = 0$
$\begin{aligned}
\text{Equation of lines: } & 3x - 2y - 1 = 0 \;\;\; \cdots \; (1) \\\\
& 4x + y - 27 = 0 \;\;\; \cdots \; (2)
\end{aligned}$
Multiplying equation $(2)$ by $2$ and adding to equation $(1)$ gives
$11x - 55 = 0$ $\implies$ $x = 5$
Substituting the value of $x$ in equation $(1)$ gives
$15 - 2y -1 = 0$ $\implies$ $y = 7$
$\therefore$ $\;$ The point of intersection of lines $(1)$ and $(2)$ is $P \left(5,7\right)$
Given: Center of the circle $= C \left(2,3\right)$
Radius of the circle $= CP = \sqrt{\left(5 - 2\right)^2 + \left(7 - 3\right)^2} = 5$
$\therefore$ $\;$ Equation of circle with center $\left(2,3\right)$ and radius $5$ is
$\left(x - 2\right)^2 + \left(y - 3\right)^2 = 5^2$
i.e. $x^2 + y^2 - 4x - 6y - 12 = 0$