Find the equation of the circle which passes through the points $\left(\dfrac{1}{3}, \dfrac{2}{3}\right)$, $\left(\dfrac{-1}{3}, \dfrac{2}{3}\right)$ and $\left(\dfrac{4}{3}, \dfrac{5}{3}\right)$.
Let the equation of the required circle be $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$
Since the required circle passes through the points $\left(\dfrac{1}{3}, \dfrac{2}{3}\right)$, $\left(\dfrac{-1}{3}, \dfrac{2}{3}\right)$ and $\left(\dfrac{4}{3}, \dfrac{5}{3}\right)$, we have
$\left(\dfrac{1}{3}\right)^2 + \left(\dfrac{2}{3}\right)^2 + 2 \times \dfrac{1}{3} \times g + 2 \times \dfrac{2}{3} \times f + c = 0$
i.e. $\;$ $\dfrac{1}{9} + \dfrac{4}{9} + \dfrac{2}{3}g + \dfrac{4}{3}f + c = 0$
i.e. $\;$ $5 + 6g + 12f + 9c = 0$ $\;\;\; \cdots \; (1)$
$\left(\dfrac{-1}{3}\right)^2 + \left(\dfrac{2}{3}\right)^2 + 2 \times \dfrac{-1}{3} \times g + 2 \times \dfrac{2}{3} \times f + c = 0$
i.e. $\;$ $\dfrac{1}{9} + \dfrac{4}{9} - \dfrac{2}{3}g + \dfrac{4}{3}f + c = 0$
i.e. $\;$ $5 - 6g + 12f + 9c = 0$ $\;\;\; \cdots \; (2)$
$\left(\dfrac{4}{3}\right)^2 + \left(\dfrac{5}{3}\right)^2 + 2 \times \dfrac{4}{3} \times g + 2 \times \dfrac{5}{3} \times f + c = 0$
i.e. $\;$ $\dfrac{16}{9} + \dfrac{25}{9} + \dfrac{8}{3}g + \dfrac{10}{3}f + c = 0$
i.e. $\;$ $41 + 24g + 30f + 9c = 0$ $\;\;\; \cdots \; (3)$
Subtracting equation $(1)$ from equation $(2)$ we have, $\;\;\;$ $g = 0$
Substituting $g = 0$ in equations $(1)$ and $(3)$ we have,
$5 + 12f + 9c = 0$ $\;\;\; \cdots \; (4)$
$41 + 30f + 9c = 0$ $\;\;\; \cdots \; (5)$
Solving equations $(4)$ and $(5)$ simultaneously gives
$f = -2$ $\;\;$ and $\;$ $c = \dfrac{19}{9}$
$\therefore \;$ The equation of the required circle is
$x^2 + y^2 + 2 \times 0 \; x + 2 \times \left(-2\right)y + \dfrac{19}{9} = 0$
i.e. $\;$ $9 x^2 + 9 y^2 - 36y + 19 = 0$