1. **Problem:** Find the equation of the tangent line to the circle $$x^2 + y^2 + 14x + 18y - 39 = 0$$ at the point in the second quadrant where $$x = -2$$. 2. **Rewrite the circle equation in standard form:** Complete the square for $$x$$ and $$y$$ terms: $$x^2 + 14x + y^2 + 18y = 39$$ For $$x$$: $$x^2 + 14x = (x + 7)^2 - 49$$ For $$y$$: $$y^2 + 18y = (y + 9)^2 - 81$$ So the equation becomes: $$(x + 7)^2 - 49 + (y + 9)^2 - 81 = 39$$ Simplify: $$(x + 7)^2 + (y + 9)^2 = 39 + 49 + 81 = 169$$ This is a circle centered at $$(-7, -9)$$ with radius $$13$$. 3. **Find the point on the circle where $$x = -2$$ in the second quadrant:** Substitute $$x = -2$$: $$(-2 + 7)^2 + (y + 9)^2 = 169$$ $$5^2 + (y + 9)^2 = 169$$ $$25 + (y + 9)^2 = 169$$ $$(y + 9)^2 = 144$$ $$y + 9 = \\pm 12$$ So, $$y = 3$$ or $$y = -21$$. Since the point is in the second quadrant, $$x < 0$$ and $$y > 0$$, so $$y = 3$$. The point is $$(-2, 3)$$. 4. **Find the slope of the radius at this point:** The radius connects center $$(-7, -9)$$ to point $$(-2, 3)$$. Slope of radius: $$m_r = \frac{3 - (-9)}{-2 - (-7)} = \frac{12}{5}$$ 5. **Find the slope of the tangent line:** Tangent is perpendicular to radius, so $$m_t = -\frac{1}{m_r} = -\frac{1}{\frac{12}{5}} = -\frac{5}{12}$$ 6. **Write the equation of the tangent line using point-slope form:** $$y - y_1 = m_t (x - x_1)$$ $$y - 3 = -\frac{5}{12}(x + 2)$$ Multiply both sides by 12: $$12(y - 3) = -5(x + 2)$$ $$12y - 36 = -5x - 10$$ Bring all terms to one side: $$5x + 12y - 26 = 0$$ 7. **Answer:** The equation of the tangent line is $$5x + 12y - 26 = 0$$. This corresponds to option c.