Find the coordinates of the centre of the circle \(3x^{2}+3y^{2} – 4x + 8y -2=0\)

Explanation

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

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

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