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

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

Correct Answer: Option C

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})