1. **State the problem:** Find the equation of the tangent line to the circle given by $$x^2 + y^2 + 14x + 18y - 39 = 0$$ at the point in the second quadrant where $$x = -2$$. 2. **Rewrite the circle equation in standard form:** Complete the square for both $$x$$ and $$y$$. $$x^2 + 14x + y^2 + 18y = 39$$ For $$x$$: $$x^2 + 14x = (x + 7)^2 - 49$$ For $$y$$: $$y^2 + 18y = (y + 9)^2 - 81$$ Substitute back: $$ (x + 7)^2 - 49 + (y + 9)^2 - 81 = 39 $$ Simplify: $$ (x + 7)^2 + (y + 9)^2 = 39 + 49 + 81 = 169 $$ So the circle has center $$C(-7, -9)$$ and radius $$r = \sqrt{169} = 13$$. 3. **Find the point on the circle where $$x = -2$$ in the second quadrant:** Substitute $$x = -2$$ into the circle equation: $$ (-2 + 7)^2 + (y + 9)^2 = 169 $$ Calculate: $$ 5^2 + (y + 9)^2 = 169 $$ $$ 25 + (y + 9)^2 = 169 $$ $$ (y + 9)^2 = 144 $$ So: $$ y + 9 = \pm 12 $$ Two possible $$y$$ values: - $$y = 3$$ - $$y = -21$$ Since the point is in the second quadrant where $$x < 0$$ and $$y > 0$$, choose $$y = 3$$. So the point of tangency is $$P(-2, 3)$$. 4. **Find the slope of the radius from center to point:** $$ m_{radius} = \frac{3 - (-9)}{-2 - (-7)} = \frac{12}{5} $$ 5. **Find the slope of the tangent line:** The tangent line is perpendicular to the radius, so its slope is the negative reciprocal: $$ m_{tangent} = -\frac{5}{12} $$ 6. **Write the equation of the tangent line using point-slope form:** $$ y - 3 = -\frac{5}{12}(x + 2) $$ Multiply both sides by 12 to clear denominator: $$ 12(y - 3) = -5(x + 2) $$ Simplify: $$ 12y - 36 = -5x - 10 $$ Bring all terms to one side: $$ 5x + 12y - 26 = 0 $$ 7. **Final answer:** The equation of the tangent line is $$\boxed{5x + 12y - 26 = 0}$$, which corresponds to option c.