B4. A circle with radius $r=4$ has its center at one of the vertices of the ellipse $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$ and passes through both foci and the other vertex. Write the equation of the ellipse $(a>b)$!

## Solutions to problems

## First year

I. PART

| 1 | 2 | 3 | 4 | 5 | 6 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| $\mathrm{D}$ | $\mathrm{D}$ | $\mathrm{E}$ | $\mathrm{E}$ | $\mathrm{A}$ | $\mathrm{B}$ |
B4. We see that the center of the circle must lie at $(0, b)$ or $(0, -b)$. Due to symmetry, we can take the first point. Since the foci are on the x-axis, the circle must pass through the vertices on the y-axis, so $r=2b$ or $b=2$.

The sum of the distances from any point on the ellipse to the foci is equal to $2a$. Therefore, $2a=2r$ or $a=4$. The equation of the ellipse is $4x^{2}+16y^{2}=4 \cdot 16$ or $x^{2}+4y^{2}=16$.