WAEC Further Mathematics 2017 Paper

The equation for a circle with centre coordinates (a, b) and radius r is

$(x-a)^{2} + (y-b)^{2} = r^{2}$

Expanding the above equation, we have

$x^{2} - 2ax +a^{2} + y^{2} - 2by + b^{2} - r^{2} = 0$ so that

$x^{2} - 2ax + y^{2} - 2by = r^{2} - a^{2} - b^{2}$

Taking the original equation given, $3x^{2} + 3y^{2} - 4x + 8y = 2$ and making the coefficients of $x^{2}$ and $y^{2}$ = 1,

$x^{2} + y^{2} - \;\frac{4x}{3} + \;\frac{8y}{3} = \;\frac{2}{3}$, comparing, we have

$2a = \;\frac{4}{3}; 2b = \;\frac{-8}{3}$

$\implies a = \;\frac{2}{3}; b = \;\frac{-4}{3}$