1. **State the problem:** We have a circle defined by the equation $$x^2 + y^2 = 25$$. 2. **Find the centre and radius:** The general form of a circle equation is $$ (x - h)^2 + (y - k)^2 = r^2 $$ where $ (h, k) $ is the centre and $ r $ is the radius. 3. **Identify centre and radius:** Here, the equation is $$ x^2 + y^2 = 25 $$ which can be rewritten as $$ (x - 0)^2 + (y - 0)^2 = 5^2 $$. So, the centre is $$ (0, 0) $$ and the radius is $$ 5 $$. 4. **Show that the point (3, -4) lies on the circle:** Substitute $x=3$ and $y=-4$ into the circle equation: $$ 3^2 + (-4)^2 = 9 + 16 = 25 $$ Since the left side equals the right side, the point lies on the circle. 5. **Find two other points on the circle:** Choose values for $x$ and solve for $y$ or vice versa. For example, let $x=0$: $$ 0^2 + y^2 = 25 \Rightarrow y^2 = 25 \Rightarrow y = \pm 5 $$ Points: $$ (0, 5) $$ and $$ (0, -5) $$. Alternatively, let $y=0$: $$ x^2 + 0^2 = 25 \Rightarrow x^2 = 25 \Rightarrow x = \pm 5 $$ Points: $$ (5, 0) $$ and $$ (-5, 0) $$. **Final answers:** - Centre: $$ (0, 0) $$ - Radius: $$ 5 $$ - Point (3, -4) lies on the circle. - Two other points on the circle: $$ (0, 5) $$ and $$ (0, -5) $$ (or $$ (5, 0) $$ and $$ (-5, 0) $$).