6. A circle, the center of which lies on the line $y=b$, intersects the parabola $y=\frac{3}{4} x^{2}$ at least at three points; one of these points is the origin, and two of the remaining points lie on the line $y=\frac{3}{4} x+b$. Find all values of $b$ for which the described configuration is possible.
Answer: $b=\frac{25}{12}$.

Solution. First, consider $b>0$. Let the origin be denoted as $O(0 ; 0)$, the center of the circle as $Q(a ; b)$ (since it lies on the line $y=b$, its ordinate is $b$); the points of intersection of the line with the parabola as $A\left(x_{1} ; y_{1}\right)$ and $B\left(x_{2} ; y_{2}\right)\left(x_{1}0\right)$. Also, let $T(0 ; b)$ be the point of intersection of the given line with the ordinate axis, and $C$ be the point of intersection of the circle with the ordinate axis, different from $O$.

Triangle $Q O C$ is isosceles ($Q O=Q C$ as radii), and $Q T$ is its height, hence $Q T$ is also a median, $C T=O T$, so the point $C$ has coordinates $(0 ; 2 b)$. Drop a perpendicular $A H$ from point $A$ to the ordinate axis. Then $\angle T A H$ is the angle of inclination of the line, its tangent is $\frac{3}{4}$. Therefore, $\cos \angle T A H=\frac{4}{5}, A T=A H: \cos \angle T A H=\frac{5}{4} A H=-\frac{5 x_{1}}{4}$. Similarly, we find that $B T=\frac{5 x_{2}}{4}$.

$A B$ and $O C$ are two chords of the given circle. By the intersecting chords theorem, ${ }^{1} C T \cdot O T=$ $A T \cdot B T$, i.e., $b \cdot b=-\frac{5 x_{1}}{4} \cdot \frac{5 x_{2}}{4}$. The abscissas $x_{1}$ and $x_{2}$ of the points of intersection of the line $y=\frac{3}{4} x+b$ and the parabola $y=\frac{3}{4} x^{2}$ are determined by the equation $\frac{3}{4} x^{2}=\frac{3}{4} x+b \Leftrightarrow x^{2}-x-\frac{4 b}{3}=0$. By Vieta's theorem, $x_{1} x_{2}=-\frac{4 b}{3}$. Thus, $b^{2}=-\frac{25}{16} \cdot\left(-\frac{4 b}{3}\right) \Leftrightarrow b^{2}=\frac{25 b}{12}$, from which $b=\frac{25}{12}$.

The value $b=0$ is not suitable, as in this case the given line takes the form $y=\frac{3}{4} x$, i.e., passes through the origin.

For $b<0$ (naturally, we consider only those $b$ for which the line and the parabola have two points of intersection), both numbers $x_{1}$ and $x_{2}$ are positive. Point $T$ is the midpoint of segment $O C$ (we retain all the notations from the first case). Then, on the one hand, it turns out that point $T$ is the midpoint of chord $O C$, i.e., lies inside the circle. On the other hand, points $A$ and $B$ lie on the circle, so $A B$ is a chord of this circle, and point $T$ lies on the extension of chord $A B$, i.e., outside the circle. We get a contradiction, and this case is impossible.