1. **Problem 5a:** Verify if point P(-7, -6) lies on the circle with equation $$x^2 + 18x + y^2 - 2y + 29 = 0$$. 2. Substitute $$x = -7$$ and $$y = -6$$ into the circle equation: $$(-7)^2 + 18(-7) + (-6)^2 - 2(-6) + 29 = 49 - 126 + 36 + 12 + 29$$ 3. Simplify: $$49 - 126 + 36 + 12 + 29 = (49 + 36 + 12 + 29) - 126 = 126 - 126 = 0$$ 4. Since the left side equals zero, point P lies on the circle. --- 5. **Problem 5b:** Find the equation of the tangent to the circle at point P(-7, -6). 6. First, find the center of the circle by completing the square: Rewrite the equation: $$x^2 + 18x + y^2 - 2y + 29 = 0$$ Group x and y terms: $$ (x^2 + 18x) + (y^2 - 2y) = -29$$ Complete the square: $$x^2 + 18x = (x + 9)^2 - 81$$ $$y^2 - 2y = (y - 1)^2 - 1$$ Substitute back: $$(x + 9)^2 - 81 + (y - 1)^2 - 1 = -29$$ Simplify: $$(x + 9)^2 + (y - 1)^2 = 53$$ Center is $$(-9, 1)$$ and radius $$r = \sqrt{53}$$. 7. The radius vector at P is from center to P: $$\vec{r} = (-7 + 9, -6 - 1) = (2, -7)$$ 8. The slope of radius is: $$m_r = \frac{-7}{2}$$ 9. The tangent line is perpendicular to radius, so its slope is: $$m_t = -\frac{1}{m_r} = -\frac{1}{-7/2} = \frac{2}{7}$$ 10. Equation of tangent line at P(-7, -6): $$y - (-6) = \frac{2}{7}(x - (-7))$$ $$y + 6 = \frac{2}{7}(x + 7)$$ Simplify: $$y = \frac{2}{7}x + 2 - 6 = \frac{2}{7}x - 4$$ --- 11. **Problem 6a:** Find coordinates of S and T where tangent to circle $$(x + 4)^2 + (y - 1)^2 = 242$$ at point (7, -10) meets y-axis and x-axis. 12. Center is $$(-4, 1)$$, radius $$r = \sqrt{242}$$. 13. Slope of radius to point (7, -10): $$m_r = \frac{-10 - 1}{7 + 4} = \frac{-11}{11} = -1$$ 14. Tangent slope is perpendicular: $$m_t = -\frac{1}{m_r} = -\frac{1}{-1} = 1$$ 15. Equation of tangent line at (7, -10): $$y - (-10) = 1(x - 7)$$ $$y + 10 = x - 7$$ $$y = x - 17$$ 16. Find S (y-intercept): set $$x=0$$ $$y = 0 - 17 = -17$$ So, $$S = (0, -17)$$. 17. Find T (x-intercept): set $$y=0$$ $$0 = x - 17 \Rightarrow x = 17$$ So, $$T = (17, 0)$$. --- 18. **Problem 6b:** Find area of triangle OST where O is origin, S(0, -17), T(17, 0). 19. Area formula for triangle with vertices at O(0,0), S(0,-17), T(17,0): $$\text{Area} = \frac{1}{2} |x_1 y_2 - x_2 y_1| = \frac{1}{2} |0 \times 0 - 17 \times (-17)| = \frac{1}{2} (289) = 144.5$$ --- 20. **Problem 7:** Circle $$ (x + 5)^2 + (y + 3)^2 = 80 $$, line l tangent with gradient 2. 21. Equation of line l: $$y = 2x + c$$ 22. Substitute into circle: $$(x + 5)^2 + (2x + c + 3)^2 = 80$$ 23. Expand: $$x^2 + 10x + 25 + 4x^2 + 4x(c + 3) + (c + 3)^2 = 80$$ 24. Combine terms: $$5x^2 + x(10 + 4c + 12) + 25 + (c + 3)^2 - 80 = 0$$ Simplify linear term: $$5x^2 + x(4c + 22) + (c + 3)^2 - 55 = 0$$ 25. For tangency, discriminant $$D=0$$: $$D = (4c + 22)^2 - 4 \times 5 \times ((c + 3)^2 - 55) = 0$$ 26. Expand and simplify: $$(4c + 22)^2 - 20((c + 3)^2 - 55) = 0$$ $$(16c^2 + 176c + 484) - 20(c^2 + 6c + 9 - 55) = 0$$ $$(16c^2 + 176c + 484) - 20(c^2 + 6c - 46) = 0$$ $$(16c^2 + 176c + 484) - 20c^2 - 120c + 920 = 0$$ $$-4c^2 + 56c + 1404 = 0$$ Divide by -4: $$c^2 - 14c - 351 = 0$$ 27. Solve quadratic: $$c = \frac{14 \pm \sqrt{196 + 1404}}{2} = \frac{14 \pm \sqrt{1600}}{2} = \frac{14 \pm 40}{2}$$ 28. Two values: $$c_1 = \frac{14 + 40}{2} = 27$$ $$c_2 = \frac{14 - 40}{2} = -13$$ 29. Equations of tangent lines: $$y = 2x + 27$$ $$y = 2x - 13$$ --- 30. **Problem 8a:** Line $$2x + y - 5 = 0$$ tangent to circle $$(x - 3)^2 + (y - p)^2 = 5$$. 31. Distance from center $$(3, p)$$ to line equals radius $$r = \sqrt{5}$$. 32. Distance formula: $$d = \frac{|2(3) + 1(p) - 5|}{\sqrt{2^2 + 1^2}} = \frac{|6 + p - 5|}{\sqrt{5}} = \frac{|p + 1|}{\sqrt{5}}$$ 33. Set equal to radius: $$\frac{|p + 1|}{\sqrt{5}} = \sqrt{5} \Rightarrow |p + 1| = 5$$ 34. Solve: $$p + 1 = 5 \Rightarrow p = 4$$ $$p + 1 = -5 \Rightarrow p = -6$$ --- 35. **Problem 8b:** Centers: $$ (3, 4) \text{ and } (3, -6) $$ --- 36. **Problem 9a:** Circle center $$P(11, -5)$$ passing through $$Q(5, 3)$$. 37. Radius: $$r = \sqrt{(5 - 11)^2 + (3 + 5)^2} = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$ 38. Equation: $$(x - 11)^2 + (y + 5)^2 = 100$$ --- 39. **Problem 9b:** Equation of tangent line $$l_1$$ at $$Q(5, 3)$$. 40. Slope of radius: $$m_r = \frac{3 + 5}{5 - 11} = \frac{8}{-6} = -\frac{4}{3}$$ 41. Tangent slope: $$m_t = -\frac{1}{m_r} = \frac{3}{4}$$ 42. Equation: $$y - 3 = \frac{3}{4}(x - 5)$$ $$y = \frac{3}{4}x - \frac{15}{4} + 3 = \frac{3}{4}x - \frac{15}{4} + \frac{12}{4} = \frac{3}{4}x - \frac{3}{4}$$ --- 43. **Problem 9c:** Line $$l_2$$ parallel to $$l_1$$ through midpoint of $$PQ$$. 44. Midpoint: $$M = \left(\frac{11 + 5}{2}, \frac{-5 + 3}{2}\right) = (8, -1)$$ 45. Equation of $$l_2$$: $$y - (-1) = \frac{3}{4}(x - 8)$$ $$y + 1 = \frac{3}{4}x - 6$$ $$y = \frac{3}{4}x - 7$$ 46. Find intersections with circle: $$(x - 11)^2 + (y + 5)^2 = 100$$ Substitute $$y = \frac{3}{4}x - 7$$: $$(x - 11)^2 + \left(\frac{3}{4}x - 7 + 5\right)^2 = 100$$ $$(x - 11)^2 + \left(\frac{3}{4}x - 2\right)^2 = 100$$ 47. Expand: $$(x^2 - 22x + 121) + \left(\frac{9}{16}x^2 - 3x + 4\right) = 100$$ 48. Combine: $$x^2 - 22x + 121 + \frac{9}{16}x^2 - 3x + 4 = 100$$ $$\left(1 + \frac{9}{16}\right)x^2 - 25x + 125 = 100$$ $$\frac{25}{16}x^2 - 25x + 125 = 100$$ 49. Simplify: $$\frac{25}{16}x^2 - 25x + 25 = 0$$ Divide by 25: $$\frac{1}{16}x^2 - x + 1 = 0$$ Multiply by 16: $$x^2 - 16x + 16 = 0$$ 50. Solve quadratic: $$x = \frac{16 \pm \sqrt{256 - 64}}{2} = \frac{16 \pm \sqrt{192}}{2} = \frac{16 \pm 8\sqrt{3}}{2} = 8 \pm 4\sqrt{3}$$ 51. Corresponding y values: $$y = \frac{3}{4}x - 7$$ For $$x = 8 + 4\sqrt{3}$$: $$y = \frac{3}{4}(8 + 4\sqrt{3}) - 7 = 6 + 3\sqrt{3} - 7 = -1 + 3\sqrt{3}$$ For $$x = 8 - 4\sqrt{3}$$: $$y = \frac{3}{4}(8 - 4\sqrt{3}) - 7 = 6 - 3\sqrt{3} - 7 = -1 - 3\sqrt{3}$$ 52. Points: $$A = \left(8 + 4\sqrt{3}, -1 + 3\sqrt{3}\right), B = \left(8 - 4\sqrt{3}, -1 - 3\sqrt{3}\right)$$ --- 53. **Problem 9d:** Length of segment $$AB$$. 54. Use distance formula: $$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ $$= \sqrt{\left((8 - 4\sqrt{3}) - (8 + 4\sqrt{3})\right)^2 + \left((-1 - 3\sqrt{3}) - (-1 + 3\sqrt{3})\right)^2}$$ $$= \sqrt{(-8\sqrt{3})^2 + (-6\sqrt{3})^2} = \sqrt{64 \times 3 + 36 \times 3} = \sqrt{192 + 108} = \sqrt{300}$$ Simplify: $$\sqrt{300} = 10\sqrt{3}$$ --- 55. **Problem 10a:** Points $$R(2, 3)$$ and $$S(10, 1)$$ on circle with center $$C(a, -2)$$. Midpoint $$M$$ of $$RS$$: $$M = \left(\frac{2 + 10}{2}, \frac{3 + 1}{2}\right) = (6, 2)$$ Line $$l$$ passes through $$M$$ and $$C$$. Slope of $$l$$: $$m = \frac{-2 - 2}{a - 6} = \frac{-4}{a - 6}$$ Equation of $$l$$: $$y - 2 = \frac{-4}{a - 6}(x - 6)$$ --- 56. **Problem 10b:** Find $$a$$. Since $$R$$ lies on circle: $$(2 - a)^2 + (3 + 2)^2 = r^2$$ Similarly, $$S$$ lies on circle: $$(10 - a)^2 + (1 + 2)^2 = r^2$$ Equate: $$(2 - a)^2 + 25 = (10 - a)^2 + 9$$ Expand: $$4 - 4a + a^2 + 25 = 100 - 20a + a^2 + 9$$ Simplify: $$29 - 4a = 109 - 20a$$ $$-4a + 20a = 109 - 29$$ $$16a = 80 \Rightarrow a = 5$$ --- 57. **Problem 10c:** Equation of circle with center $$(5, -2)$$. Radius squared: $$(2 - 5)^2 + (3 + 2)^2 = (-3)^2 + 5^2 = 9 + 25 = 34$$ Equation: $$(x - 5)^2 + (y + 2)^2 = 34$$ --- 58. **Problem 10d:** Find points of intersection $$A$$ and $$B$$ of line $$l$$ and circle. Line $$l$$ with $$a=5$$: $$y - 2 = \frac{-4}{5 - 6}(x - 6) = \frac{-4}{-1}(x - 6) = 4(x - 6)$$ $$y = 4x - 24 + 2 = 4x - 22$$ Substitute into circle: $$(x - 5)^2 + (4x - 22 + 2)^2 = 34$$ Simplify: $$(x - 5)^2 + (4x - 20)^2 = 34$$ Expand: $$x^2 - 10x + 25 + 16x^2 - 160x + 400 = 34$$ Combine: $$17x^2 - 170x + 425 = 34$$ $$17x^2 - 170x + 391 = 0$$ Divide by 17: $$x^2 - 10x + 23 = 0$$ Solve quadratic: $$x = \frac{10 \pm \sqrt{100 - 92}}{2} = \frac{10 \pm \sqrt{8}}{2} = 5 \pm \sqrt{2}$$ Corresponding y: $$y = 4x - 22$$ For $$x = 5 + \sqrt{2}$$: $$y = 4(5 + \sqrt{2}) - 22 = 20 + 4\sqrt{2} - 22 = -2 + 4\sqrt{2}$$ For $$x = 5 - \sqrt{2}$$: $$y = 4(5 - \sqrt{2}) - 22 = 20 - 4\sqrt{2} - 22 = -2 - 4\sqrt{2}$$ Points: $$A = (5 + \sqrt{2}, -2 + 4\sqrt{2}), B = (5 - \sqrt{2}, -2 - 4\sqrt{2})$$