Find the equation of the circle which passes through the points $\left(6,1\right)$, $\left(4,3\right)$ and $\left(7,9\right)$.
Let the equation of the required circle be $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$
Since the required circle passes through the points $\left(6,1\right)$, $\left(4,3\right)$ and $\left(7,9\right)$, we have
$6^2 + 1^2 + 2 \times 6 \times g + 2 \times 1 \times f + c = 0$ $\;\;$ i.e. $\;$ $37 + 12g + 2f + c = 0$ $\;\;\; \cdots \; (1)$
$4^2 + 3^2 + 2 \times 4 \times g + 2 \times 3 \times f + c = 0$ $\;\;$ i.e. $\;$ $25 + 8 g + 6f + c = 0$ $\;\;\; \cdots \; (2)$
$7^2 + 9^2 + 2 \times 7 \times g + 2 \times 9 \times f + c = 0$ $\;\;$ i.e. $\;$ $130 + 14 g + 18 f + c = 0$ $\;\;\; \cdots \; (3)$
Subtracting equation $(1)$ from equation $(2)$ we have,
$-4g + 4f - 12 = 0$ $\;\;$ i.e. $\;$ $g - f = -3$ $\;\;\; \cdots \; (4)$
Subtracting equation $(2)$ from equation $(3)$ gives
$6g + 12f + 105 = 0$ $\;\;$ i.e. $\;$ $2g + 4f = -35$ $\;\;\; \cdots \; (5)$
Solving equations $(4)$ and $(5)$ simultaneously gives
$g = \dfrac{-47}{6}$ $\;\;$ and $\;$ $f = \dfrac{-29}{6}$
Substituting the values of $f$ and $g$ in equation $(1)$ gives
$37 + 12 \times \left(\dfrac{-47}{6}\right) + 2 \times \left(\dfrac{-29}{6}\right) + c = 0$ $\implies$ $c = \dfrac{200}{3}$
$\therefore \;$ The equation of the required circle is
$x^2 + y^2 + 2 \times \left(\dfrac{-47}{6}\right) x + 2 \times \left(\dfrac{-29}{6}\right)y + \dfrac{200}{3} = 0$
i.e. $\;$ $3 x^2 + 3 y^2 - 47x - 29y + 200 = 0$