10.2. Given the coefficients $a, b, c$ of the quadratic trinomial $a x^{2}+b x+c$. Its graph intersects the coordinate axes at three points, and a circle passing through these points intersects the y-axis at one more point. Find the ordinate of this fourth point.
Answer: $1 / a$. Solution. First, consider the case when the parabola intersects the $O x$ axis at points $x_{1}, x_{2}$ on the same side of the origin $O$. If $a>0$, then $c>0$ and applying the theorem of intersecting chords for the circle, we get that the desired ordinate $y_{0}$ satisfies the equation $y_{0} \cdot c=x_{1} x_{2}$. But by Vieta's theorem, $x_{1} x_{2}=\frac{c}{a}$, from which we obtain $y_{0}=\frac{1}{a}$. If $a <0$, taking into account the negative signs of $c$ and $y_{0}$, the property of intersecting chords can be written as $\left(-y_{0}\right) \cdot(-c)=x_{1} x_{2}$ and we again obtain the same result $y_{0}=\frac{1}{a}$. If the roots have different signs, then instead of the theorem of intersecting chords, we should apply the theorem of intersecting secants, and then we similarly obtain the same formula for $y_{0}$.