The circle Chas equation \[ x^{2}+y^{2}-10 x+6 y+30=0 \] Find the \( y \)-coordinates of the points where the circle \( C \) crosses the line with equation \( x=4 \), giving your answers as simplified surds. (3 marks)

To find the points where the circle \( C \) crosses the line \( x = 4 \), we will substitute \( x = 4 \) into the equation of the circle. Starting with the equation of the circle: \[ x^{2} + y^{2} - 10x + 6y + 30 = 0 \] Substituting \( x = 4 \): \[ (4)^{2} + y^{2} - 10(4) + 6y + 30 = 0 \] Calculating \( (4)^{2} = 16 \) and \( -10(4) = -40 \): \[ 16 + y^{2} - 40 + 6y + 30 = 0 \] Now, simplify: \[ y^{2} + 6y + 6 = 0 \] Next, we can solve this quadratic equation using the quadratic formula \( y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \), where \( a = 1, b = 6, c = 6 \). Calculating the discriminant: \[ b^{2} - 4ac = 6^{2} - 4(1)(6) = 36 - 24 = 12 \] Now substituting into the quadratic formula: \[ y = \frac{-6 \pm \sqrt{12}}{2(1)} = \frac{-6 \pm 2\sqrt{3}}{2} \] This simplifies to: \[ y = -3 \pm \sqrt{3} \] Thus, the \( y \)-coordinates where the circle crosses the line \( x = 4 \) are: \[ y = -3 + \sqrt{3} \quad \text{and} \quad y = -3 - \sqrt{3} \]