For what value of $\lambda$ does there exist a pair of values $x, y$ that satisfy the following system of equations:

$$
x^{2}+y^{2}=8 x+6 y ; \quad 9 x^{2}+y^{2}=6 y ; \quad y^{2}+9=\lambda x+6 y+\lambda
$$

What is the geometric meaning of this problem?
Let's transform our equations into the following form:

$$
\begin{aligned}
& y^{2}-6 y=8 x-x^{2} \\
& y^{2}-6 y=-9 x^{2} \\
& y^{2}-6 y=\lambda(x+1)-9
\end{aligned}
$$

The solutions to the system of equations (1) and (2) are to be determined, and we will examine for which values of $\lambda$ these solutions satisfy equation (3).

By equating the right-hand sides of (1) and (2):

$$
8 x^{2}+8 x=0, \quad \text { or } \quad x(x+1)=0
$$

from which

$$
x_{1}=0, \quad x_{2}=-1
$$

Substituting $x_{1}$ into (2):

$$
y(y-6)=0
$$

from which

$$
y_{1,1}=0, \quad y_{1,2}=6
$$

Substituting $x_{2}$:

$$
y^{2}-6 y+9=0, \quad \text { or } \quad(y-3)^{2}=0
$$

from which

$$
y_{2,1}=y_{2,2}=3
$$

Thus, the solutions to (1) and (2) are $(0,0)$; $(0,6)$; and the double-counted $(-1,3)$.

Substituting these pairs into (3):

$$
\begin{array}{rlrl}
0 & =\lambda-9, & \text { from which } \quad \lambda=9 \\
36-36 & =\lambda-9, & & \text { from which } \lambda=9 \\
0 & =0, & & \text { which is an identity independent of } \lambda.
\end{array}
$$

The first two pairs satisfy all three equations when $\lambda=9$, while the third pair satisfies them for any value of $\lambda$. Since we have considered all solutions to the system (1) and (2), there can be no more solutions to the system.

For the geometric interpretation of the problem and solution, let's transform the equations.

In (1), move all terms to the left side and add $3^{2}+4^{2}$ to both sides:

$$
\begin{gathered}
x^{2}-8 x+16+y^{2}-6 y+9=25 \\
(x-4)^{2}+(y-3)^{2}=5^{2}
\end{gathered}
$$

This is the equation of a circle with a radius of 5 units and center at $(4,3)$.

Similarly, transforming (2):

$$
\frac{x^{2}}{1^{2}}+\frac{(y-3)^{2}}{3^{2}}=1^{2}
$$

This is the equation of an ellipse with center at $(0,3)$, semi-major axis of 3 units, semi-minor axis of 1 unit, and the major axis along the $y$-axis of the coordinate system.

The transformed form of (3) is:

$$
(y-3)^{2}=\lambda(x+1)
$$

This represents a parabola. Its axis is parallel to the $x$-axis but shifted 3 units up along the $y$-axis. The common vertex of the parabolas is at the point $(-1,3)$. The parameter $\lambda$ affects the spread of the parabola's branches and determines whether the parabolas are located to the right of $x=-1$ (for $\lambda>0$) or to the left (for $\lambda<0$). For $\lambda=0$, we get a degenerate parabola, which is the line $y=3$.

The problem was to determine the value of $\lambda$ for which the three curves have a common point.

The first step is to find the common points of the circle and the ellipse. We calculated that they intersect at the points $(0,0)$ and $(0,6)$, and they are tangent at the point $(-1,3)$. (See the diagram).

![](https://cdn.mathpix.com/cropped/2024_05_02_a0ad98f34b1bb03e9279g-2.jpg?height=851&width=968&top_left_y=122&top_left_x=557)

The point $(-1,3)$ is also the common vertex of the parabola family, so it is always a common point of the three curves regardless of the choice of $\lambda$.

Since the line parallel to the minor axis at $y=3$ is the symmetry axis of all three curves, and the points $(0,0)$ and $(0,6)$ are symmetric with respect to this line, if we choose $\lambda$ such that the parabola passes through one of these points, it will also pass through the other; this leads to $\lambda=9$. (The parabola corresponding to $\lambda=9$ is not shown on the diagram due to technical reasons.)

Iván Tatár (Debrecen, Ref. g. IV. o. t.)

Antal Rockenbauer (Bp. X., I. László g. IV. o. t.)