Find the coordinates of the centre of the circle \(4x^{2} + 4y^{2} - 5x + 3y - 2 = 0\).

A.

\((\frac{-5}{4}, \frac{3}{4})\)

B.

\((\frac{3}{8}, -\frac{5}{8})\)

C.

\((\frac{5}{8}, -\frac{3}{8})\)

D.

\((\frac{5}{4}, -\frac{3}{4})\)

Correct answer is C

Equation : \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Expanding : \(x^{2} + y^{2} - 2ax - 2by + a^{2} + b^{2} = r^{2}\)

Given, \(4x^{2} + 4y^{2} - 5x + 3y - 2 = 0\)

Divide through by 4 to make the coefficient of \(x^{2}\) and \(y^{2}\) to be 1.

\(x^{2} + y^{2} - \frac{5}{4}x + \frac{3}{4}y - \frac{1}{2} = 0\)

Comparing, \(2a = \frac{5}{4} \implies a = \frac{5}{8}\)

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

\((a, b) = (\frac{5}{8}, -\frac{5}{8})\)