The equation of a circle is \( x^{2}+y^{2}-6 x+4 y=k \). (1) Determine the coordinates of the centre of the circle. (2) If the radius of the circle is 13 units, determine the value of \( k \).

To solve this problem, we need to complete the square for both the \( x \) and \( y \) terms to put the equation of the circle into its standard form. The standard form of the equation of a circle is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center of the circle and \(r\) is the radius. (1) Determine the coordinates of the center of the circle. Given equation: \( x^{2}+y^{2}-6 x+4 y=k \) To complete the square for \(x\) and \(y\), we need to add and subtract the square of half the coefficient of \(x\) and \(y\) to the equation. For \(x\), the coefficient is \(-6\), so half of that is \(-3\), and the square of \(-3\) is \(9\). For \(y\), the coefficient is \(4\), so half of that is \(2\), and the square of \(2\) is \(4\). Now, we add these squares to both sides of the equation: \( x^{2} - 6x + 9 + y^{2} + 4y + 4 = k + 9 + 4 \) This simplifies to: \( (x - 3)^2 + (y + 2)^2 = k + 13 \) Now the equation is in standard form, and we can see that the center of the circle is at \((h, k)\), which is \((3, -2)\). (2) If the radius of the circle is 13 units, determine the value of \( k \). We know that the radius \(r\) is 13 units, so we can square this to get \(r^2 = 13^2 = 169\). Now we can set \(r^2\) equal to \(k + 13\) from the standard form of the circle's equation: \( 169 = k + 13 \) To solve for \(k\), we subtract 13 from both sides: \( k = 169 - 13 \) \( k = 156 \) So the value of \(k\) is 156.