Problem (Geometry):
A circle passes through the origin and the points of intersection of the parabolas $y=x^{2}-3$ and $y=-x^{2}-2 x+9$. Determine the coordinates of the centre of this circle.

Solution:
We first determine the three points through which the circle passes.

The first point is the origin $(0,0)$.

The second and third points are found by determining the points of intersection of the two parabolas $y=x^{2}-3$ and $y=-x^{2}-2 x+9$. We do this by setting the $y$ values equal.

$$
x^{2}-3=-x^{2}-2 x+9
$$

$2 x^{2}+2 x-12=0$

$x^{2}+x-6=0$

$(x+3)(x-2)=0$

so $x=-3$ or $x=2$.



We determine the points of intersection by substituting into the first parabola.

If $x=2, y=2^{2}-3=1$, so the point of intersection is $(2,1)$.

If $x=-3, y=(-3)^{2}-3=6$, so the point of intersection is $(-3,6)$.

Therefore, the circle passes through the three points $A(0,0), B(2,1)$ and $C(-3,6)$.

Let the centre of the circle be the point $Q(a, b)$.



Finding the centre of the circle can be done in a variety of ways.

We use the fact $Q$ is of equal distance from each of the points $A, B$ and $C$. In particular $Q A^{2}=Q B^{2}=Q C^{2}$ or $x^{2}+y^{2}=(x-2)^{2}+(y-1)^{2}=(x+3)^{2}+(y-6)^{2}$

From the first equality,

$$
\begin{aligned}
& x^{2}+y^{2}=(x-2)^{2}+(y-1)^{2} \\
& 4 x+2 y=5
\end{aligned}
$$





From the second equality,

$$
\begin{aligned}
(x-2)^{2}+(y-1)^{2} & =(x+3)^{2}+(y-6)^{2} \\
-10 x+10 y & =40 \\
y & =x+4
\end{aligned}
$$

Substituting the equation above into into $4 x+2 y=5$, we obtain $4 x+2(x+4)=5$ or $6 x=-3$ or $x=-\frac{1}{2}$. Thus, $y=-\frac{1}{2}+4=\frac{7}{2}$, and so the centre of the circle is $\left(-\frac{1}{2}, \frac{7}{2}\right)$.

Final answer: $(-\frac{1}{2}, \frac{7}{2})$