The circle \( C \) has equation \[ x^{2}+y^{2}-10 x+6 y+30=0 \] Find the coordinates of the centre of \( c \).

To find the center of the circle given by the equation \( x^{2}+y^{2}-10 x+6 y+30=0 \), we start by rearranging it into the standard form of a circle's equation, which is \( (x - h)^2 + (y - k)^2 = r^2 \). First, we rewrite the equation by grouping \( x \) and \( y \) terms: \[ x^{2} - 10x + y^{2} + 6y + 30 = 0 \] Now, we complete the square for the \( x \) and \( y \) terms. For \( x^{2} - 10x \): - Take half of \(-10\) which is \(-5\) and square it to get \(25\). - Rewrite as: \[ (x - 5)^{2} - 25 \] For \( y^{2} + 6y \): - Take half of \(6\) which is \(3\) and square it to get \(9\). - Rewrite as: \[ (y + 3)^{2} - 9 \] Substituting these back into the equation gives: \[ (x - 5)^{2} - 25 + (y + 3)^{2} - 9 + 30 = 0 \] Combining constants: \[ (x - 5)^{2} + (y + 3)^{2} - 4 = 0 \] Moving \( -4 \) to the other side: \[ (x - 5)^{2} + (y + 3)^{2} = 4 \] This indicates that the center of the circle is at the coordinates \( (h, k) = (5, -3) \). Thus, the coordinates of the center of circle \( C \) are \( \boxed{(5, -3)} \).