A beautiful geometry problem 
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 $ax^2 + 2bx + c = 0$.
 A: 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.) 
A: $MX=x*\dfrac x{\sqrt{d^2+x^2}}$
,$MQ=\sqrt{x^2-\dfrac{x^4}{d^2+x^2}}=\dfrac{xd}{\sqrt{d^2+x^2}}$
$PM=\sqrt{d^2-MQ^2}=d\sqrt{1-\dfrac{x^2}{d^2+x^2}}=\dfrac{d^2}{\sqrt{d^2+x^2}}$
Since $PMP'\sim MXQ'$
$p:q+x=PM:MX=d^2/\sqrt{d^2+x^2}:x^2/\sqrt{d^2+x^2}=d^2:x^2$
$⇔d^2(q+x)=px^2$
$⇔px^2-d^2x-d^2q=0$
$x=(d^2+d\sqrt{d^2+4pq})/2p $
Similarly, since $PNP'\sim NQ'Y$
$p:q-y=PN:NY=d^2:y^2 $
$py^2+d^2y-d^2q=0 $
$y=(-d^2+d\sqrt{d^2+4pq})/2p$
Then
$(X-x)(X-y)=X^2-\sqrt{d^2+4pq}/p*X+qd^2/p$.
but $(X-x)(X+y)=X^2+\dfrac{d^2}pX-\dfrac{qd^2}p$. x and y are solutions of this equation.
A: HINTS:
Direct ( BFI :) ) method 
Let radius of circle be $ r$
Transversal through circle center
$$ \frac{y-r}{x-p} = \frac{2r}{p+q} $$
Central circle Equation
$$ x^2+y^2 =r^2$$
Points of intersection $$ ( N,M)= (x_1,y_1), (x_2,y_2) $$
Form Left slant line  Equation  L in which put  $ \rightarrow y=0 $
Form Right  slant line   Equation R  in which put$ \rightarrow y=0 $
(Next verify the problem beauty etc. in hindsight.) 
