9 The line \( y=m x+1 \) intersects the circle \( x^{2}+y^{2}-19 x-51=0 \) at the point \( P(5,11) \). a Find the coordinates of the point \( Q \) where the line meets the curve again. b Find the equation of the perpendicular bisector of the line \( P Q \). c Find the \( x \)-coordinates of the points where this perpendicular bisector intersects the circle. Give your answers in exact form.

To solve the problem, we will follow the steps outlined in the prompt. ### Step 1: Extract Known Conditions We have: - The line equation: \( y = mx + 1 \) - The circle equation: \( x^2 + y^2 - 19x - 51 = 0 \) - The point of intersection \( P(5, 11) \) ### Step 2: Part (a) - Find the coordinates of the point \( Q \) 1. **Substituting the line equation into the circle equation:** We substitute \( y = mx + 1 \) into the circle equation: \[ x^2 + (mx + 1)^2 - 19x - 51 = 0 \] Expanding this: \[ x^2 + (m^2x^2 + 2mx + 1) - 19x - 51 = 0 \] Combining like terms: \[ (1 + m^2)x^2 + (2m - 19)x + (1 - 51) = 0 \] Simplifying: \[ (1 + m^2)x^2 + (2m - 19)x - 50 = 0 \] 2. **Using the point \( P(5, 11) \) to find \( m \):** Since \( P \) lies on the line, we can substitute \( x = 5 \) and \( y = 11 \): \[ 11 = m(5) + 1 \implies 10 = 5m \implies m = 2 \] 3. **Substituting \( m \) back into the quadratic equation:** Now substituting \( m = 2 \): \[ (1 + 2^2)x^2 + (2(2) - 19)x - 50 = 0 \] This simplifies to: \[ 5x^2 - 15x - 50 = 0 \] Dividing through by 5: \[ x^2 - 3x - 10 = 0 \] 4. **Solving the quadratic equation:** Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-10)}}{2(1)} = \frac{3 \pm \sqrt{9 + 40}}{2} = \frac{3 \pm 7}{2} \] This gives: \[ x = 5 \quad \text{or} \quad x = -2 \] Since \( P(5, 11) \) is one intersection, the other point \( Q \) has \( x = -2 \). 5. **Finding the \( y \)-coordinate of \( Q \):** Substitute \( x = -2 \) into the line equation: \[ y = 2(-2) + 1 = -4 + 1 = -3 \] Thus, the coordinates of point \( Q \) are \( Q(-2, -3) \). ### Step 3: Part (b) - Find the equation of the perpendicular bisector of line \( PQ \) 1. **Finding the midpoint \( M \) of \( PQ \):** The midpoint \( M \) is given by: \[ M = \left( \frac{5 + (-2)}{2}, \frac{11 + (-3)}{2} \right) = \left( \frac{3}{2}, 4 \right) \] 2. **Finding the slope of line \( PQ \):** The slope \( m_{PQ} \) is: \[ m_{PQ} = \frac{-3 - 11}{-2 - 5} = \frac{-14}{-7} = 2 \] The slope of the perpendicular bisector \( m_{perpendicular} \) is: \[ m_{perpendicular} = -\frac{1}{m_{PQ}} = -\frac{1}{2} \] 3. **Using point-slope form to find the equation:** The equation of the perpendicular bisector is: \[ y - 4 = -\frac{1}{2}\left(x - \frac{3}{2}\right) \] Simplifying: \[ y - 4 = -\frac{1}{2}x + \frac{3}{4} \] \[ y = -\frac{1}{2}x + \frac{3}{4} + 4 = -\frac{1}{2}x + \frac{19}{4} \] ### Step 4: Part (c) - Find the \( x \)-coordinates of the points where the perpendicular bisector intersects the circle 1. **Substituting the perpendicular bisector equation into the circle equation:** Substitute \( y = -\frac{1}{2}x + \frac{19}{4} \) into the circle equation: \[ x^2 + \left(-\frac{1}{2}x + \frac{19}{4}\right)^2 - 19x - 51 = 0 \] Expanding: \[ x^2 + \left(\frac{1}{4}x^2 - \frac{19}{4}x + \frac{361}{16}\right) - 19x - 51 = 0 \] Combining like terms: \[ \left(1 + \frac{1}{4}\right)x^2 + \left(-\frac{19}{4} - 19\right)x + \left(\frac{361}{16} - 51\right) = 0 \] This simplifies to: \[ \frac{5}{4}x^2 - \frac{95}{4}x + \left(\frac{361 - 816}{16}\right) = 0 \] \[ \frac{5}{4}x^2 - \frac{95}{4}x - \frac{455}{16} = 0 \] Multiplying through by 16 to eliminate fractions: \[ 20x^2 - 380x - 455 = 0 \] 2. **Using the quadratic formula to find \( x \):** \[ x = \frac{-(-380) \pm \sqrt{(-380)^2 - 4(20)(-455)}}{2(20)} \]