1. **Problem:** Find the coordinates of the centre and the radius of the circle given by the equation $$9x^2 + 9y^2 + 6x - 24y + 8 = 0$$. 2. **Step 1: Simplify the equation.** Divide the entire equation by 9 to get the standard form: $$x^2 + y^2 + \frac{6}{9}x - \frac{24}{9}y + \frac{8}{9} = 0$$ which simplifies to $$x^2 + y^2 + \frac{2}{3}x - \frac{8}{3}y + \frac{8}{9} = 0$$. 3. **Step 2: Rearrange terms to complete the square.** Group $x$ and $y$ terms: $$x^2 + \frac{2}{3}x + y^2 - \frac{8}{3}y = -\frac{8}{9}$$. 4. **Step 3: Complete the square for $x$ and $y$.** - For $x$: Take half of $\frac{2}{3}$, which is $\frac{1}{3}$, square it to get $\left(\frac{1}{3}\right)^2 = \frac{1}{9}$. - For $y$: Take half of $-\frac{8}{3}$, which is $-\frac{4}{3}$, square it to get $\left(-\frac{4}{3}\right)^2 = \frac{16}{9}$. Add these squares to both sides: $$x^2 + \frac{2}{3}x + \frac{1}{9} + y^2 - \frac{8}{3}y + \frac{16}{9} = -\frac{8}{9} + \frac{1}{9} + \frac{16}{9}$$ 5. **Step 4: Write as perfect squares and simplify right side:** $$\left(x + \frac{1}{3}\right)^2 + \left(y - \frac{4}{3}\right)^2 = \frac{9}{9} = 1$$ 6. **Step 5: Identify centre and radius.** - Centre: $\left(-\frac{1}{3}, \frac{4}{3}\right)$ - Radius: $\sqrt{1} = 1$ **Final answer:** The centre is $\left(-\frac{1}{3}, \frac{4}{3}\right)$ and the radius is $1$.