<s>[INST] Condition for a line to intersect a curve The line $lx + my + n = 0$ intersects the curve $ax^2+ 2hxy + by^2 = 1$ at the point $P$ and $Q$. The circle on $PQ$ as diameter passes through the origin.
We have to prove that $$n^2(a^2 + b^2) = l^2+ m^2$$
I write the equation of circle as $C+\lambda L=0$
But satisfying $(0,0)$ I got $\lambda =\dfrac{1}{n}$
Now how to proceed?
 [/INST] Let the circle passes through $O(0,0)$, $P(x_1,y_1)$ and $Q(x_2,y_2)$.
Now,
$$
\left \{
  \begin{align*}
    aX^2+2hXY+bY^2 &= 1 \\
    \ell X+mY +n &= 0
  \end{align*}
\right.$$
$$\implies
aX^2-2hX\left( \frac{\ell X+n}{m} \right)+
b\left( \frac{\ell X+n}{m} \right)^2=1$$
$$\left( a-\frac{2h\ell}{m}+\frac{b\ell^2}{m^2}  \right)X^2+
2n\left( \frac{b\ell}{m^2}-\frac{h}{m} \right)X+\frac{bn^2}{m^2}-1=0$$
$$x_1 x_2=\frac{bn^2-m^2}{am^2-2h\ell m+bl^2}$$
Similarly,
$$\left( b-\frac{2hm}{\ell}+\frac{am^2}{\ell^2} \right)Y^2+
2n\left( \frac{am}{\ell^2}-\frac{h}{\ell} \right)Y+\frac{an^2}{\ell^2}-1=0$$
$$y_1 y_2=\frac{an^2-\ell^2}{am^2-2h\ell m+b\ell^2}$$
Since $PQ$ is the diameter, $OP \perp OQ$.

We have
  $$\frac{y_1}{x_1} \times \frac{y_2}{x_2}=-1$$

$$\frac{an^2-\ell^2}{bn^2-m^2}=-1$$
$$\fbox{$n^2(a+b)=\ell^2+m^2$}$$
</s>