1. **State the problem:** We have a circle given by the equation $$x^2 + y^2 + 6x - 2y - 26 = 0$$ and a line $$y = kx - 5$$. We want to find the values of the constant $k$ such that the line intersects the circle at two distinct points. 2. **Rewrite the circle equation in standard form:** Complete the square for $x$ and $y$. For $x$: $$x^2 + 6x = x^2 + 6x + 9 - 9 = (x+3)^2 - 9$$ For $y$: $$y^2 - 2y = y^2 - 2y + 1 - 1 = (y-1)^2 - 1$$ Substitute back: $$ (x+3)^2 - 9 + (y-1)^2 - 1 - 26 = 0 $$ $$ (x+3)^2 + (y-1)^2 - 36 = 0 $$ $$ (x+3)^2 + (y-1)^2 = 36 $$ This is a circle with center $(-3,1)$ and radius $6$. 3. **Substitute the line equation into the circle:** Replace $y$ by $kx - 5$ in the circle equation: $$ (x+3)^2 + (kx - 5 - 1)^2 = 36 $$ $$ (x+3)^2 + (kx - 6)^2 = 36 $$ Expand: $$ (x+3)^2 = x^2 + 6x + 9 $$ $$ (kx - 6)^2 = k^2 x^2 - 12kx + 36 $$ Sum: $$ x^2 + 6x + 9 + k^2 x^2 - 12kx + 36 = 36 $$ Simplify: $$ x^2 + k^2 x^2 + 6x - 12kx + 9 + 36 - 36 = 0 $$ $$ (1 + k^2) x^2 + (6 - 12k) x + 9 = 0 $$ 4. **Condition for two distinct intersection points:** The quadratic in $x$ must have two distinct real roots, so its discriminant $\Delta$ must be positive: $$ \Delta = b^2 - 4ac > 0 $$ where $$ a = 1 + k^2, \quad b = 6 - 12k, \quad c = 9 $$ Calculate: $$ \Delta = (6 - 12k)^2 - 4(1 + k^2)(9) $$ $$ = 36 - 144k + 144k^2 - 36 - 36k^2 $$ $$ = 144k^2 - 36k^2 - 144k + 36 - 36 $$ $$ = 108k^2 - 144k $$ 5. **Solve inequality:** $$ 108k^2 - 144k > 0 $$ Divide both sides by 12: $$ 9k^2 - 12k > 0 $$ Factor: $$ 3k(3k - 4) > 0 $$ 6. **Analyze sign:** The product $3k(3k - 4)$ is positive when both factors are positive or both are negative. - Case 1: $3k > 0$ and $3k - 4 > 0$ \Rightarrow $k > 0$ and $k > \frac{4}{3}$ \Rightarrow $k > \frac{4}{3}$ - Case 2: $3k < 0$ and $3k - 4 < 0$ \Rightarrow $k < 0$ and $k < \frac{4}{3}$ (always true) \Rightarrow $k < 0$ 7. **Final answer:** $$ \boxed{k < 0 \text{ or } k > \frac{4}{3}} $$ These are the values of $k$ for which the line intersects the circle at two distinct points.