1. **State the problem:** We have a circle $C$ with equation $$x^2 + y^2 - 2x + 6y - 30 = 0.$$ Points $P(3,3)$ and $Q(7,-5)$ lie on $C$. We need to: a) Find the tangent line to $C$ at $Q$ in the form $ax + by + c = 0$. b) Find the perpendicular bisector of segment $PQ$. c) Find the intersection of the tangent and the perpendicular bisector. --- 2. **Rewrite the circle equation in center-radius form:** Complete the square for $x$ and $y$: $$x^2 - 2x + y^2 + 6y = 30$$ For $x$: $$x^2 - 2x = (x - 1)^2 - 1$$ For $y$: $$y^2 + 6y = (y + 3)^2 - 9$$ Substitute back: $$(x - 1)^2 - 1 + (y + 3)^2 - 9 = 30$$ $$(x - 1)^2 + (y + 3)^2 = 30 + 1 + 9 = 40$$ So the circle has center $C(1, -3)$ and radius $r = \sqrt{40} = 2\sqrt{10}$. --- 3. **Part (a): Find the tangent line at $Q(7,-5)$** The radius vector from center to $Q$ is: $$\overrightarrow{CQ} = (7 - 1, -5 + 3) = (6, -2)$$ The tangent line at $Q$ is perpendicular to $\overrightarrow{CQ}$. If the radius vector has slope: $$m_r = \frac{-2}{6} = -\frac{1}{3}$$ Then the tangent slope $m_t$ is the negative reciprocal: $$m_t = 3$$ Using point-slope form for tangent at $Q(7,-5)$: $$y - (-5) = 3(x - 7)$$ $$y + 5 = 3x - 21$$ $$y = 3x - 26$$ Rewrite in standard form: $$3x - y - 26 = 0$$ --- 4. **Part (b): Find the perpendicular bisector of $PQ$** Points: $$P(3,3), Q(7,-5)$$ Midpoint $M$: $$M = \left( \frac{3+7}{2}, \frac{3 + (-5)}{2} \right) = (5, -1)$$ Slope of $PQ$: $$m_{PQ} = \frac{-5 - 3}{7 - 3} = \frac{-8}{4} = -2$$ Slope of perpendicular bisector is negative reciprocal: $$m_{pb} = \frac{1}{2}$$ Equation of perpendicular bisector through $M(5,-1)$: $$y - (-1) = \frac{1}{2}(x - 5)$$ $$y + 1 = \frac{1}{2}x - \frac{5}{2}$$ $$y = \frac{1}{2}x - \frac{7}{2}$$ Multiply both sides by 2 to clear fractions: $$2y = x - 7$$ Rewrite in standard form: $$x - 2y - 7 = 0$$ --- 5. **Part (c): Find intersection of tangent and perpendicular bisector** Solve system: $$\begin{cases} 3x - y - 26 = 0 \\ x - 2y - 7 = 0 \end{cases}$$ From second equation: $$x = 2y + 7$$ Substitute into first: $$3(2y + 7) - y - 26 = 0$$ $$6y + 21 - y - 26 = 0$$ $$5y - 5 = 0$$ $$5y = 5$$ $$y = 1$$ Find $x$: $$x = 2(1) + 7 = 9$$ So intersection point is: $$(9, 1)$$ --- **Final answers:** a) Tangent line at $Q$: $$3x - y - 26 = 0$$ b) Perpendicular bisector of $PQ$: $$x - 2y - 7 = 0$$ c) Intersection point: $$(9, 1)$$