1. **State the problem:** Find the equation of the normal to the circle given by $$x^2 + y^2 - 2x - 4y - 12 = 0$$ at the point $$(-3, 5)$$. 2. **Rewrite the circle equation in standard form:** Group terms: $$x^2 - 2x + y^2 - 4y = 12$$ Complete the square for $$x$$ and $$y$$: $$x^2 - 2x + 1 + y^2 - 4y + 4 = 12 + 1 + 4$$ $$ (x - 1)^2 + (y - 2)^2 = 17 $$ So the circle has center $$C(1, 2)$$ and radius $$ adius = \\sqrt{17}$$. 3. **Find the slope of the radius at point $$(-3, 5)$$:** Slope of radius $$m_r = \frac{5 - 2}{-3 - 1} = \frac{3}{-4} = -\frac{3}{4}$$. 4. **Find the slope of the tangent line:** The tangent is perpendicular to the radius, so $$m_t = -\frac{1}{m_r} = -\frac{1}{-3/4} = \frac{4}{3}$$. 5. **Find the slope of the normal line:** The normal line is along the radius, so its slope is the same as the radius: $$m_n = m_r = -\frac{3}{4}$$. 6. **Write the equation of the normal line passing through $$(-3, 5)$$:** Using point-slope form: $$y - 5 = -\frac{3}{4}(x + 3)$$ Multiply both sides by 4: $$4(y - 5) = -3(x + 3)$$ Expand: $$4y - 20 = -3x - 9$$ Bring all terms to one side: $$3x + 4y - 11 = 0$$ 7. **Compare with given options:** None of the options exactly match $$3x + 4y - 11 = 0$$. Check if any option is a multiple or rearrangement: - a. $$-3x - 6y - 2 = 0$$ - b. $$x - 2y - 8 = 0$$ - c. $$5x + 6y - 3 = 0$$ - d. $$-4y - 3x + 8 = 0$$ which can be rewritten as $$-3x - 4y + 8 = 0$$ or $$3x + 4y - 8 = 0$$ Option d is close but constant term differs. Since the point $$(-3, 5)$$ must satisfy the normal line equation, test options: For option d: $$-4(5) - 3(-3) + 8 = -20 + 9 + 8 = -3 \neq 0$$ For option a: $$-3(-3) - 6(5) - 2 = 9 - 30 - 2 = -23 \neq 0$$ For option b: $$(-3) - 2(5) - 8 = -3 - 10 - 8 = -21 \neq 0$$ For option c: $$5(-3) + 6(5) - 3 = -15 + 30 - 3 = 12 \neq 0$$ None satisfy the point. **Conclusion:** The correct normal line equation is $$3x + 4y - 11 = 0$$ which is not among the options. **Final answer:** Equation of the normal line is $$3x + 4y - 11 = 0$$.