Find the center and radius of the circle whose equation is $\;$ $x^2 + y^2 + \dfrac{4}{7}x - \dfrac{6}{7}y = \dfrac{36}{49}$
Given equation of circle: $\;$ $x^2 = y^2 + \dfrac{4}{7}x - \dfrac{6}{7}y = \dfrac{36}{49}$
i.e. $\;$ $x^2 + y^2 + \dfrac{4}{7}x - \dfrac{6}{7}y - \dfrac{36}{49} = 0$
Comparing with the standard equation of a circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives
$2g = \dfrac{4}{7}$ $\implies$ $g = \dfrac{2}{7}$, $\;$ $2f = \dfrac{-6}{7}$ $\implies$ $f = \dfrac{-3}{7}$, $\;$ $c = \dfrac{-36}{49}$
Center of circle $= \left(-g, -f\right) = \left(\dfrac{-2}{7}, \dfrac{3}{7}\right)$
Radius of circle $= r = \sqrt{g^2 + f^2 - c} = \sqrt{\left(\dfrac{2}{7}\right)^2 + \left(\dfrac{-3}{7}\right)^2 + \dfrac{36}{49}}$
i.e. $\;$ $r = \sqrt{\dfrac{4}{49} + \dfrac{9}{49} + \dfrac{36}{49}} = \sqrt{\dfrac{49}{49}} = 1$