<s>[INST] 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 $ax^2 + 2bx + c = 0$.
 [/INST] $MX=x*\dfrac x{\sqrt{d^2+x^2}}$
,$MQ=\sqrt{x^2-\dfrac{x^4}{d^2+x^2}}=\dfrac{xd}{\sqrt{d^2+x^2}}$
$PM=\sqrt{d^2-MQ^2}=d\sqrt{1-\dfrac{x^2}{d^2+x^2}}=\dfrac{d^2}{\sqrt{d^2+x^2}}$
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.
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