1. **State the problem:** We have a circle given by $$x^2 + y^2 - 4x - 2y - 69 = 0$$ and a line $$2y = mx + 15$$ that intersects the circle at point $$P(9,6)$$. We need to: a) Find the other intersection point $$Q$$ of the line and circle. b) Find the equation of the perpendicular bisector of segment $$PQ$$. c) Find the exact $$x$$-coordinates where this perpendicular bisector intersects the circle. 2. **Rewrite the line equation:** From $$2y = mx + 15$$, we get $$y = \frac{mx}{2} + \frac{15}{2}$$. 3. **Use point P to find $$m$$:** Since $$P(9,6)$$ lies on the line, $$6 = \frac{m \times 9}{2} + \frac{15}{2}$$ Multiply both sides by 2: $$12 = 9m + 15$$ $$9m = 12 - 15 = -3$$ $$m = -\frac{1}{3}$$ 4. **Substitute $$m$$ back into the line equation:** $$y = -\frac{1}{3} \times \frac{x}{2} + \frac{15}{2} = -\frac{x}{6} + \frac{15}{2}$$ 5. **Find point Q by solving the system:** Substitute $$y = -\frac{x}{6} + \frac{15}{2}$$ into the circle equation: $$x^2 + y^2 - 4x - 2y - 69 = 0$$ Replace $$y$$: $$x^2 + \left(-\frac{x}{6} + \frac{15}{2}\right)^2 - 4x - 2\left(-\frac{x}{6} + \frac{15}{2}\right) - 69 = 0$$ Expand and simplify: $$x^2 + \left(\frac{x^2}{36} - 2.5x + \frac{225}{4}\right) - 4x + \frac{x}{3} - 15 - 69 = 0$$ Combine like terms: $$x^2 + \frac{x^2}{36} - 2.5x - 4x + \frac{x}{3} + \frac{225}{4} - 15 - 69 = 0$$ Calculate coefficients: $$\left(1 + \frac{1}{36}\right)x^2 + \left(-2.5 - 4 + \frac{1}{3}\right)x + \left(\frac{225}{4} - 84\right) = 0$$ $$\frac{37}{36}x^2 + \left(-6.5 + 0.3333\right)x + \left(56.25 - 84\right) = 0$$ $$\frac{37}{36}x^2 - 6.1667x - 27.75 = 0$$ Multiply entire equation by 36 to clear denominators: $$37x^2 - 222x - 999 = 0$$ 6. **Solve quadratic for $$x$$:** Use quadratic formula: $$x = \frac{222 \pm \sqrt{222^2 - 4 \times 37 \times (-999)}}{2 \times 37}$$ Calculate discriminant: $$222^2 = 49284$$ $$4 \times 37 \times 999 = 147852$$ $$\Delta = 49284 + 147852 = 197136$$ Simplify $$\sqrt{197136}$$: $$197136 = 36 \times 5476$$ $$\sqrt{197136} = 6 \sqrt{5476}$$ Factor 5476: $$5476 = 4 \times 1369$$ $$\sqrt{5476} = 2 \times 37 = 74$$ So, $$\sqrt{197136} = 6 \times 74 = 444$$ Therefore, $$x = \frac{222 \pm 444}{74}$$ Two solutions: $$x_1 = \frac{222 + 444}{74} = \frac{666}{74} = 9$$ (point P) $$x_2 = \frac{222 - 444}{74} = \frac{-222}{74} = -3$$ 7. **Find $$y$$ for point Q:** $$y = -\frac{x}{6} + \frac{15}{2} = -\frac{-3}{6} + \frac{15}{2} = \frac{1}{2} + \frac{15}{2} = 8$$ So, $$Q(-3, 8)$$. 8. **Find midpoint M of $$PQ$$:** $$M_x = \frac{9 + (-3)}{2} = 3$$ $$M_y = \frac{6 + 8}{2} = 7$$ 9. **Find slope of $$PQ$$:** $$m_{PQ} = \frac{8 - 6}{-3 - 9} = \frac{2}{-12} = -\frac{1}{6}$$ 10. **Slope of perpendicular bisector:** $$m_{perp} = -\frac{1}{m_{PQ}} = 6$$ 11. **Equation of perpendicular bisector:** Using point-slope form with midpoint $$M(3,7)$$: $$y - 7 = 6(x - 3)$$ $$y = 6x - 18 + 7 = 6x - 11$$ 12. **Find intersection of perpendicular bisector and circle:** Substitute $$y = 6x - 11$$ into circle: $$x^2 + (6x - 11)^2 - 4x - 2(6x - 11) - 69 = 0$$ Expand: $$x^2 + 36x^2 - 132x + 121 - 4x - 12x + 22 - 69 = 0$$ Combine terms: $$37x^2 - 148x + 74 = 0$$ 13. **Solve quadratic for $$x$$:** $$x = \frac{148 \pm \sqrt{(-148)^2 - 4 \times 37 \times 74}}{2 \times 37}$$ Calculate discriminant: $$148^2 = 21904$$ $$4 \times 37 \times 74 = 10952$$ $$\Delta = 21904 - 10952 = 10952$$ Simplify $$\sqrt{10952}$$: $$10952 = 4 \times 2738$$ $$\sqrt{10952} = 2 \sqrt{2738}$$ 14. **Final exact $$x$$-coordinates:** $$x = \frac{148 \pm 2\sqrt{2738}}{74} = \frac{74 \pm \sqrt{2738}}{37}$$ --- **Final answers:** - a) $$Q(-3, 8)$$ - b) Perpendicular bisector: $$y = 6x - 11$$ - c) $$x$$-coordinates of intersections: $$x = \frac{74 \pm \sqrt{2738}}{37}$$