For what values of $a, b$ does the system of equations

$$
19 x^{2}+19 y^{2}+a x+b y+98=0 \quad \text { and } \quad 98 x^{2}+98 y^{2}+a x+b y+19=0
$$

have a common solution?
Starting from the second equation, we get

$$
79 x^{2}+79 y^{2}-79=0
$$

which simplifies to

$$
x^{2}+y^{2}=1
$$

Now, let's add the original equations:

$$
117 x^{2}+117 y^{2}+2 a x+2 b y+117=0
$$

Substituting (1) into this, we get

$$
a x+b y=-117
$$

The original equations will have a common solution if and only if (1) and (2) do.

First, let's consider the case when $a=0$.

In this case, from (2) we have $b \neq 0$, so $y=-\frac{117}{b}$. From (1), however, $y^{2} \leq 1$, so $b^{2} \geq 117^{2}$ must hold ($b \geq 117$ or $b \leq-117$), and in this case (1) and (2) will always have a common solution.

In the case where $a \neq 0$, from (2) we have $x=\frac{-117-b y}{a}$, and substituting this into (1), then multiplying by $a^{2}(>0)$ and rearranging, we get the quadratic equation

$$
\left(a^{2}+b^{2}\right) y^{2}+(2 \cdot 117 \cdot b) y+\left(117^{2}-a^{2}\right)=0
$$

The leading coefficient of this equation, $a^{2}+b^{2}>0$. This equation will have a solution if and only if its discriminant is greater than or equal to zero, that is,

$$
4\left(117^{2} b^{2}-\left(a^{2}+b^{2}\right)\left(117^{2}-a^{2}\right) \geq 0\right.
$$

Rearranging this inequality and dividing by $a^{2}>0$, we get

$$
a^{2}+b^{2} \geq 117^{2}
$$

This condition includes the case when $a=0$, so (1) and (2), and thus the original system of equations, will have a solution if and only if (3) holds, that is, for all pairs $(a, b)$ in the plane that lie on or outside the circle centered at the origin with a radius of 117 units.

Based on the work of Tóth Ágnes (Hajdúszoboszló, Hőgyes E. Gymnasium, 9th grade)

Note. With some knowledge of coordinate geometry, it can be seen that the solvability of the system of equations (1) and (2) means that the line given by equation (2) intersects or is tangent to the unit circle centered at the origin.

The distance of the line from the origin is $\frac{117}{\sqrt{a^{2}+b^{2}}}$, so we get a common point if and only if this distance is not greater than 1, the radius of the circle, that is, $117^{2} \leq a^{2}+b^{2}$.