1. **State the problem:** We are given the circle equation $$x^2 + y^2 + 6x - 2y - 26 = 0$$ and the line equation $$y = kx - 5$$. We need to find the values of $$k$$ for which the line intersects the circle at two distinct points. 2. **Rewrite the circle equation in standard form:** Group $$x$$ and $$y$$ terms: $$x^2 + 6x + y^2 - 2y = 26$$ Complete the square for $$x$$ and $$y$$: $$x^2 + 6x = (x + 3)^2 - 9$$ $$y^2 - 2y = (y - 1)^2 - 1$$ Substitute back: $$(x + 3)^2 - 9 + (y - 1)^2 - 1 = 26$$ Simplify: $$(x + 3)^2 + (y - 1)^2 = 26 + 9 + 1 = 36$$ So the circle has center $$(-3,1)$$ and radius $$6$$. 3. **Substitute the line equation into the circle equation:** Replace $$y$$ with $$kx - 5$$: $$(x + 3)^2 + ((kx - 5) - 1)^2 = 36$$ Simplify inside the second square: $$(x + 3)^2 + (kx - 6)^2 = 36$$ 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: $$(1 + k^2) x^2 + (6 - 12k) x + 9 + 36 - 36 = 0$$ $$(1 + k^2) x^2 + (6 - 12k) x + 9 = 0$$ 4. **Analyze the quadratic in $$x$$:** For the line to intersect the circle at two distinct points, the quadratic must have two distinct real roots. The discriminant $$\Delta$$ must be positive: $$\Delta = b^2 - 4ac > 0$$ Here, $$a = 1 + k^2$$, $$b = 6 - 12k$$, $$c = 9$$. Calculate $$\Delta$$: $$\Delta = (6 - 12k)^2 - 4(1 + k^2)(9)$$ $$= 36 - 144k + 144k^2 - 36 - 36k^2$$ $$= 108k^2 - 144k$$ 5. **Solve the inequality:** $$108k^2 - 144k > 0$$ Divide both sides by 12: $$9k^2 - 12k > 0$$ Factor: $$3k(3k - 4) > 0$$ 6. **Determine intervals for $$k$$:** The product $$3k(3k - 4) > 0$$ means both factors are positive or both are negative. - Case 1: $$3k > 0$$ and $$3k - 4 > 0$$ $$\Rightarrow k > 0$$ and $$k > \frac{4}{3}$$ So $$k > \frac{4}{3}$$. - Case 2: $$3k < 0$$ and $$3k - 4 < 0$$ $$\Rightarrow k < 0$$ and $$k < \frac{4}{3}$$ (always true if $$k < 0$$) So $$k < 0$$. 7. **Final answer:** The line intersects the circle at two distinct points if $$k < 0 \quad \text{or} \quad k > \frac{4}{3}$$.