1. **State the problem:** Given the circle equation $$x^2 + y^2 - 4x + 6y + 4 = 0$$, find the centre and radius. 2. **Formula and rules:** The general form of a circle is $$ (x - h)^2 + (y - k)^2 = r^2 $$ where $ (h, k) $ is the centre and $ r $ is the radius. 3. **Rewrite the equation by completing the square:** Group $x$ and $y$ terms: $$ x^2 - 4x + y^2 + 6y = -4 $$ Complete the square for $x$: $$ x^2 - 4x = (x - 2)^2 - 4 $$ Complete the square for $y$: $$ y^2 + 6y = (y + 3)^2 - 9 $$ 4. **Substitute back:** $$ (x - 2)^2 - 4 + (y + 3)^2 - 9 = -4 $$ Simplify: $$ (x - 2)^2 + (y + 3)^2 - 13 = -4 $$ $$ (x - 2)^2 + (y + 3)^2 = 9 $$ 5. **Identify centre and radius:** Centre: $ (2, -3) $ Radius: $ \sqrt{9} = 3 $ 6. **Answer:** The correct choice is c. centre (2, -3), radius 3.