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 step by step, let's tackle each part of the question: **Part a: Finding the coordinates of point \( Q \)** The equation of the circle given is: \[ x^2 + y^2 - 19x - 51 = 0 \] We can rewrite this equation in standard form. Completing the square: \[ x^2 - 19x + y^2 = 51 \implies (x - \frac{19}{2})^2 + y^2 = \frac{361}{4} + 51 = \frac{361 + 204}{4} = \frac{565}{4} \] This represents a circle centered at \( \left(\frac{19}{2}, 0\right) \) with radius \( \frac{\sqrt{565}}{2} \). Now let’s use the intersection of the line \( y = mx + 1 \) with the circle. We substitute \( y \) in the circle's equation: \[ x^2 + (mx + 1)^2 - 19x - 51 = 0 \] Expanding this gives: \[ x^2 + (m^2x^2 + 2mx + 1) - 19x - 51 = 0 \implies (1 + m^2)x^2 + (2m - 19)x + (1 - 51) = 0 \] This simplifies to: \[ (1 + m^2)x^2 + (2m - 19)x - 50 = 0 \] Since point \( P(5, 11) \) lies on this line, it is one of the roots. To find the other root, we can use Vieta's formulas. Let the roots be \( x_1 = 5 \) and \( x_2 \). From Vieta's, we know \( x_1 + x_2 = -\frac{b}{a} \): \[ 5 + x_2 = \frac{19 - 2m}{1 + m^2} \implies x_2 = \frac{19 - 2m}{1 + m^2} - 5 \] Calculating \( x_2 \) will require knowing \( m \). Since the slope \( m \) can be calculated by the slope between points \( P(5, 11) \) and the center of the circle, which we will use further in our perpendicular bisector calculations. To further find \( y \) coordinate of \( Q \), substitute \( x_2 \) back into the line equation. **Part b: Finding the equation of the perpendicular bisector of line \( PQ \)** Let \( Q=(x_2, y_2) \) where \( y_2 = mx_2 + 1\). The midpoint \( M \) of \( P \) and \( Q \) is: \[ M = \left(\frac{5 + x_2}{2}, \frac{11 + y_2}{2}\right) \] The slope of line \( PQ \) is: \[ \text{slope} = \frac{y_2 - 11}{x_2 - 5} \] The slope of the perpendicular bisector is the negative reciprocal: \[ \text{slope of perpendicular} = -\frac{x_2 - 5}{y_2 - 11} \] The equation of the perpendicular bisector in point-slope form: \[ y - \frac{11 + y_2}{2} = -\frac{x_2 - 5}{y_2 - 11} \left(x - \frac{5 + x_2}{2}\right) \] This can be expanded and rearranged to get the explicit form. **Part c: Finding the x-coordinates of where the perpendicular bisector intersects the circle** Let’s denote this equation as \( f(x) = \) the equation of the perpendicular bisector in \( y \). Substitute this into the original circle's equation: \[ (x^2 + (f(x))^2 - 19x - 51 = 0) \] This results in a polynomial that can be solved for \( x \). Calculate the discriminant to ensure the intersection points indeed exist. From this setup, you'll calculate the coordinates by solving the obtained polynomial, yielding the x-coordinates.