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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 $QX = x$.

enter image description here

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 $ax^2 + 2bx + c = 0$.

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    Are $P$ and $Q$ supposed to be on the circle, as your drawing suggests? – 2017-02-11
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    Yes, I'm sorry, I'll edit the post. – 2017-02-11
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    This problem appears in a book by Felix Klein (in french, *"Leçons sur certaines questions de Géométrie Élémentaire"*, translation of the german *"Vorträge über ausgewählte Fragen der Elementargeometrie"*). The solution given uses the inversion, and proves first that $X,M,N$ and $Y$ are cocyclic. I'll eventually post a complete solution later, if nobody steps in. – 2017-02-11
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    The german original may be found [here](https://archive.org/details/vortrgeberausge00kleigoog) for instance. The problem is near p. 31 I think. An english translation, *"Famous Problems of Elementary Geometry"*, may be found [here](https://archive.org/details/famousproblemsof00kleiiala) (see p. 37). – 2017-02-11
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    I think it was p.38. – 2017-02-11
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    @TakahiroWaki Ah, you are probably right. I wasn't able to find *exactly* things in the same order as in my translation, so it was a guess. – 2017-02-11
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    I wonder if it is possible to calculate it out using power of a point, Pythagorean theorem, etc. – 2017-02-11

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

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$$\begin{align} \triangle PP^\prime M\sim \triangle XQ^\prime M &\implies \frac{|\overline{XQ^\prime}|}{|\overline{PP^\prime}|} = \frac{d\sin^2\theta}{d\cos^2\theta} \quad \left( = \tan^2\theta \right)\\[4pt] &\implies \frac{q+x}{p} = \frac{x^2}{d^2} \end{align}$$ 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.)

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    If d=2, it fits perfectly. I forgot to mention that, but anyway I was after a more general solution for the problem, so thank you very much! – 2017-02-12
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    @R.Maia I understand beautiful solutions you say, but it's near to beaitiful. It's probably depends on definition of p, q, x. – 2017-02-13
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$MX=x*\dfrac x{\sqrt{d^2+x^2}}$ ,$MQ=\sqrt{x^2-\dfrac{x^4}{d^2+x^2}}=\dfrac{xd}{\sqrt{d^2+x^2}}$ $PM=\sqrt{d^2-MQ^2}=d\sqrt{1-\dfrac{x^2}{d^2+x^2}}=\dfrac{d^2}{\sqrt{d^2+x^2}}$

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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    I follow the derivations of the equations in x and y. Very nice. But what is going on in the last two lines? What is X? Not point X in the OP's diagram of course. Please make the argument a little more explicit here. – 2017-02-13
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    @EdwardPorcella If you read OP's statement till last line, you can make sense. I tried to create beatiful equation of two solution. – 2017-02-13
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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.)