Question:

Step 1: Set up the equation of line AB passing through point A. The line $AB$ passes through point $A(-10, 0)$ with unknown slope $m$. We can write this line in the form: $$y = m(x + 10)$$ Rearranging to standard form: $$mx - y + 10m = 0$$ Step 2: Apply the tangency condition using the distance formula. For line $AB$ to be tangent to the circle $x^2 + y^2 = 9$ (which has center at the origin and radius $r = 3$), the perpendicular distance from the center $(0, 0)$ to the line must equal the radius. Using the point-to-line distance formula: $$\frac{|m(0) - 0 + 10m|}{\sqrt{m^2 + 1}} = 3$$ Simplifying the numerator: $$\frac{|10m|}{\sqrt{m^2 + 1}} = 3$$ Step 3: Solve for the slope m. Squaring both sides to eliminate the absolute value: $$\frac{100m^2}{m^2 + 1} = 9$$ Cross-multiplying: $$100m^2 = 9(m^2 + 1)$$ $$100m^2 = 9m^2 + 9$$ $$91m^2 = 9$$ $$m^2 = \frac{9}{91}$$ Taking the square root: $$m = \pm\frac{3}{\sqrt{91}} = \pm\frac{3\sqrt{91}}{91}$$ Step 4: Determine the correct sign using the position of point B. Since point $B(a, b)$ is located in the third quadrant, we have $a < 0$ and $b < 0$. Point $A$ is at $(-10, 0)$ on the negative $x$-axis. For the line to go from $A(-10, 0)$ to a point $B$ in the third quadrant (where both coordinates are negative), the line must have a negative slope as we move rightward and downward. Therefore: $$m = -\frac{3\sqrt{91}}{91}$$ The slope of $AB$ is $\boxed{-\dfrac{3\sqrt{91}}{91}}$, which corresponds to **Option 2**.