1. **State the problem:** Find which of the given lines are tangent to the circle defined by the equation $$(x - 2)^2 + (y + 5)^2 = 20.$$ 2. **Recall the formula for tangency:** A line is tangent to a circle if the distance from the center of the circle to the line equals the radius of the circle. 3. **Identify the circle's center and radius:** The circle is centered at $$(2, -5)$$ with radius $$r = \sqrt{20} = 2\sqrt{5}.$$ 4. **General form of the line:** Each line is given in slope-intercept form $$y = mx + b.$$ We will convert each to standard form $$Ax + By + C = 0$$ to use the distance formula. 5. **Distance from point to line formula:** For a point $$(x_0, y_0)$$ and line $$Ax + By + C = 0,$$ the distance $$d$$ is $$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}.$$ 6. **Check each line:** - A: $$y = 2x + 11 \Rightarrow 2x - y + 11 = 0$$ $$d = \frac{|2(2) -1(-5) + 11|}{\sqrt{2^2 + (-1)^2}} = \frac{|4 + 5 + 11|}{\sqrt{4 + 1}} = \frac{20}{\sqrt{5}} = 4\sqrt{5} \neq 2\sqrt{5}$$ (Not tangent) - B: $$y = 2x - 19 \Rightarrow 2x - y - 19 = 0$$ $$d = \frac{|2(2) -1(-5) - 19|}{\sqrt{4 + 1}} = \frac{|4 + 5 - 19|}{\sqrt{5}} = \frac{10}{\sqrt{5}} = 2\sqrt{5}$$ (Tangent) - C: $$y = -2x + 11 \Rightarrow 2x + y - 11 = 0$$ $$d = \frac{|2(2) + 1(-5) - 11|}{\sqrt{4 + 1}} = \frac{|4 - 5 - 11|}{\sqrt{5}} = \frac{12}{\sqrt{5}} \neq 2\sqrt{5}$$ (Not tangent) - D: $$y = -2x + 2 \Rightarrow 2x + y - 2 = 0$$ $$d = \frac{|2(2) + 1(-5) - 2|}{\sqrt{4 + 1}} = \frac{|4 - 5 - 2|}{\sqrt{5}} = \frac{3}{\sqrt{5}} \neq 2\sqrt{5}$$ (Not tangent) - E: $$y = 2x - 1 \Rightarrow 2x - y - 1 = 0$$ $$d = \frac{|2(2) -1(-5) - 1|}{\sqrt{4 + 1}} = \frac{|4 + 5 - 1|}{\sqrt{5}} = \frac{8}{\sqrt{5}} \neq 2\sqrt{5}$$ (Not tangent) 7. **Conclusion:** Only line B $$y = 2x - 19$$ is tangent to the circle. **Final answer:** $$\boxed{y = 2x - 19}$$