What is the coordinate of the centre of the circle \(5x^{2} + 5y^{2} - 15x + 25y - 3 = 0\)?

A.

\((\frac{15}{2}, -\frac{25}{2})\)

B.

\((\frac{3}{2}, -\frac{5}{2})\)

C.

\((-\frac{3}{2}, \frac{5}{2})\)

D.

\((-\frac{15}{2}, \frac{25}{2})\)

Correct answer is B

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

Expanding, we have:

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

Given: \(5x^{2} + 5y^{2} - 15x + 25y - 3 = 0\)

Divide through by 5,

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

Comparing, we have

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

\(-2b = 5; b = -\frac{5}{2}\)