Consider all circles in the plane whose equation can be written as

$$
\alpha\left((x-2)^{2}+y^{2}-1\right)+\beta\left((x+2)^{2}+y^{2}-1\right)=0
$$

where $\alpha$ and $\beta$ are real numbers. Where are the points that do not lie on any of these circles?
The equations $(x \pm 2)^{2}+y^{2}-1=0$ represent two circles that do not intersect or touch each other. Therefore, the coordinates of any point $P\left(x_{0}, y_{0}\right)$ in the plane satisfy at most one of these equations. Let

$$
\alpha_{0}=\left(x_{0}+2\right)^{2}+y_{0}^{2}-1 \quad \text { and } \quad \beta_{0}=-\left(\left(x_{0}-2\right)^{2}+y_{0}^{2}-1\right)
$$

Then $\alpha$ and $\beta$ are not both 0, and it is easy to see that the point $P$ lies on the figure defined by equation (1) if and only if the pair $(\alpha, \beta)$ is obtained by multiplying the pair $\left(\alpha_{0}, \beta_{0}\right)$, defined by equations (2), by some number.
![](https://cdn.mathpix.com/cropped/2024_05_02_9722f74133698a0c36d2g-1.jpg?height=404&width=812&top_left_y=470&top_left_x=645)

We also need to check for which values of $\alpha$ and $\beta$ the resulting equation will be the equation of a circle. Since the coefficients of $x^{2}$ and $y^{2}$ are the same (both are $\alpha+\beta$), if this is not 0, then the equation is either the equation of a circle or a point (the third possibility would be the empty set, e.g., $x^{2}+y^{2}+1=0$, but we know that our figure contains $P$).

If $\alpha+\beta=0$, then rearranging equation (1) gives $x=0$. Thus, the points on the $y$-axis are not on any circle whose equation is of the form (2).

If $\alpha+\beta \neq 0$, then the figure defined by (1) (a circle or a point) is symmetric with respect to the $x$-axis, because if the coordinates of point $Q\left(x_{1}, y_{1}\right)$ satisfy (1), then the coordinates of point $Q^{\prime}\left(x_{1},-y_{1}\right)$ also do. Therefore, if our figure consists of a single point, that point lies on the $x$-axis. If $\alpha=0$ (and thus $\beta \neq 0$), then (1) is the equation of a circle. If we are looking for points, we can assume that $\alpha \neq 0$, and thus that $\alpha=1$ (we can multiply our equation by $\frac{1}{\alpha}$). Rearranging the equation:

$$
(1+\beta) x^{2}+(1+\beta) y^{2}+(4 \beta-4) x+3 \beta+3=0
$$

This will be the equation of a point on the $x$-axis if and only if, after substituting $y=0$, the resulting quadratic equation in $x$ has a double root, i.e., if its discriminant is 0. This gives us:

$$
(4 \beta-4)^{2}-4(1+\beta)(3 \beta+3)=0
$$

which simplifies to $\beta=7 \pm 4 \sqrt{3}$. Substituting this back into (3) gives $x= \pm \sqrt{3}$. Therefore, the points with coordinates $( \pm \sqrt{3}, 0)$ do not lie on any circle whose equation is of the form (1).

In summary, the points on the line $x=0$ and the points $( \pm \sqrt{3}, 0)$ satisfy the conditions of the problem.

Remark. The equation in the problem represents a figure associated with a pencil of circles (for more on pencils of circles, see e.g., Hajós György: Bevezetés a geometriába, 40.8). Since the circles corresponding to the pairs $\alpha=1, \beta=0$ and $\alpha=0, \beta=1$ are disjoint, the pencil is elliptic, meaning no two distinct elements of the pencil intersect, and among the elements of the pencil, there are circles, a line, and two points. The line is the axis of symmetry of the pencil (figure).

Based on the work of Béky Bence (Fazekas M. Főv. Gyak. Gimn., 11th grade).