## Task 13/74

For which positive real numbers $a$ and $b$ does the system of equations

$$
(y-a)^{2}=1-x^{2} \quad ; \quad x^{2}=b y
$$

have exactly three real solutions?

![](https://cdn.mathpix.com/cropped/2024_06_06_1d2be4837fb839c2b31dg-247.jpg?height=668&width=625&top_left_y=631&top_left_x=264)
The equation (1) is the equation of a circle with radius 1, whose center lies on the y-axis at $a$; the equation (2) represents the equation of a parabola with its vertex at the origin (see figure). From the figure, it can be seen that exactly three solutions exist if and only if $a=r=1$ (if $a>1$, then there exist 4, 2, or no solutions; if $0<a<1$, then there are 2 solutions).

This leads to: $(y-1)^{2}=1-b y$. Solving this quadratic equation in $y$ results in $y_{1}=0, y_{2 ; 3}=2-b$ with $x_{1}=0, x_{2 ; 3}= \pm \sqrt{(2-b) b}$.

For $x_{2 ; 3}$ to be real, it is necessary and sufficient that $b \leq 2$. However, for $b=2$, $x_{2 ; 3}=0$, which must be excluded. Therefore, the system of equations has exactly three real solutions for $a=1, b<2$.