TITLE: A beautiful geometry problem
QUESTION [6 upvotes]: Let $PP'$ and $QQ'$ be two parallel lines tangent to a circle of center $C$ and radius $r$ in the points $P$ and $Q$, respectively. $P'Q'$ cuts de circle in $M$ and $N$. Let $Y$ and $X$ be the points in which $Q'Q$ is cut by $PN$ and $PM$, respectively. Given the lengths $PP'= p$, $QQ'= q$ and $2r = d$, find the lengths $QY = y$ and $QX = x$.


I've been struggling with this problem for a couple of days, so a hint or a solution would be welcome. 
Now, what makes this problem beautiful is the fact that if you let $p=-\dfrac{2a}b$ and $q =-\dfrac{c}{2b}$ then the lengths $y$ and $x$ will be the real roots of the equation $ax^2 + 2bx + c = 0$.

REPLY [3 votes]: I hadn't seen @Takahiro's argument, but mine follows the same basic approach (streamlined slightly with the help of some trig):

$$\begin{align}
\triangle PP^\prime M\sim \triangle XQ^\prime M &\implies
\frac{|\overline{XQ^\prime}|}{|\overline{PP^\prime}|} = \frac{d\sin^2\theta}{d\cos^2\theta} \quad \left( = \tan^2\theta \right)\\[4pt]
&\implies \frac{q+x}{p} = \frac{x^2}{d^2}
\end{align}$$
where $d$ is the diameter of the circle. Likewise, we have (but don't show)
$$\frac{q-y}{p} = \frac{y^2}{d^2}$$
Thus, $x$ and $-y$ (note ---as Takahiro did--- the sign change!) are roots of
$$ z^2 p - d^2 z - d^2 q = 0 \tag{$\star$}$$

Unfortunately, the substitutions $p = -\frac{2a}{b}$ and $q = -\frac{c}{2b}$ transform $(\star)$ into
$$4 a z^2 + 2 b d^2 z - c d^2 = 0 \tag{$\star\star$}$$
which is not the "beautiful" relation promised. Perhaps OP intended $p = -\frac{2 a r^2}{b} = -\frac{ad^2}{2b}$ and $q = \frac{c}{2b}$. (I'm not sure I see what's so "beautiful" about those rather ad hoc assignments, however.)