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

To find the center of the given circle, we need to write the equation in standard form, which is: \[(x - h)^2 + (y - k)^2 = r^2\] Where (h, k) is the center of the circle and r is the radius. We start by completing the square for both x and y terms, by moving the constant term to the right side and grouping the x and y terms together: \[4x^2 - 5x + 4y^2 + 3y = 2\] To complete the square for the x terms, we need to add and subtract \(\frac{5}{4}\) times the square of the coefficient of x (which is 2), inside the first bracket: \[4\left(x^2 - \frac{5}{4}x + \left(\frac{5}{4}\right)^2\right) + 4y^2 + 3y = 2 + 4\left(\frac{5}{4}\right)^2\] We do the same for the y terms, by adding and subtracting \(\frac{3}{8}\) times the square of the coefficient of y (which is \(\frac{3}{2}\)), inside the second bracket: \[4\left(x^2 - \frac{5}{4}x + \left(\frac{5}{4}\right)^2\right) + 4\left(y^2 + \frac{3}{8}y + \left(\frac{3}{16}\right)^2\right) = 2 + 4\left(\frac{5}{4}\right)^2 + 4\left(\frac{3}{16}\right)^2\] Now we have the equation in standard form, and we can read off the center and radius: Center = \(\left(\frac{5}{4}, -\frac{3}{8}\right)\) Radius = \(\sqrt{\frac{129}{16}} = \frac{3\sqrt{43}}{4}\) Therefore, the correct answer is option (C) \((\frac{5}{8}, -\frac{3}{8})\).