1. **State the problem:** We have a circle with equation $$x^2 + y^2 - 4x + 8y - 8 = 0$$. (a) Find: (i) the coordinates of the centre of the circle. (ii) the exact radius of the circle. (b) The line $$x = k$$ is tangent to the circle. Find the possible values of $$k$$. 2. **Rewrite the circle equation in standard form:** Group $$x$$ and $$y$$ terms: $$x^2 - 4x + y^2 + 8y = 8$$ Complete the square for $$x$$: $$x^2 - 4x = (x^2 - 4x + 4) - 4 = (x - 2)^2 - 4$$ Complete the square for $$y$$: $$y^2 + 8y = (y^2 + 8y + 16) - 16 = (y + 4)^2 - 16$$ Substitute back: $$(x - 2)^2 - 4 + (y + 4)^2 - 16 = 8$$ $$(x - 2)^2 + (y + 4)^2 - 20 = 8$$ $$(x - 2)^2 + (y + 4)^2 = 28$$ 3. **Find the centre and radius:** The standard form of a circle is $$(x - h)^2 + (y - k)^2 = r^2$$ Here, $$h = 2$$, $$k = -4$$, and $$r^2 = 28$$. (i) Centre: $$(2, -4)$$ (ii) Radius: $$r = \sqrt{28} = 2\sqrt{7}$$ 4. **Find the possible values of $$k$$ for the tangent line $$x = k$$:** The line $$x = k$$ is vertical. The distance from the centre to this line is the absolute difference in $$x$$-coordinates: $$d = |k - 2|$$ For the line to be tangent to the circle, the distance $$d$$ must equal the radius: $$|k - 2| = 2\sqrt{7}$$ This gives two solutions: $$k - 2 = 2\sqrt{7} \implies k = 2 + 2\sqrt{7}$$ $$k - 2 = -2\sqrt{7} \implies k = 2 - 2\sqrt{7}$$ **Final answers:** - Centre: $$(2, -4)$$ - Radius: $$2\sqrt{7}$$ - Possible values of $$k$$: $$2 + 2\sqrt{7}$$ and $$2 - 2\sqrt{7}$$