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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?

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    See this if its help http://m.meritnation.com/ask-answer/question/the-line-lx-my-n-0-intersects-the-curve-ax-2-2hxy-by-2-1-at/math/10716635 – 2017-01-30
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    If any doubt please ask. – 2017-01-30
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    @KanwaljitSingh in that condition is different, otherwise the answer is good – 2017-01-30
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    @koolman In general, $C+\lambda L=0$ is not a circle unless $a=b$ and $h=0$. – 2017-01-30

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solve the line$$lx+my+n=0$$ for $y$ we get $$y=-\frac{l}{m}x-\frac{n}{m}$$ plugging this in the equation $$ax^2+2hxy+by^2=1$$ and you will get the coordinates of $P$ and $Q$

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    How can we satisfy the condition PQ as diameter – 2017-01-30
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    the half of $PQ$ is th radius – 2017-01-30
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    have you got the intersection points $P$ and $Q$? – 2017-01-30
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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 $$\frac{y_1}{x_1} \times \frac{y_2}{x_2}=-1$$

$$\frac{an^2-\ell^2}{bn^2-m^2}=-1$$

$$\fbox{$n^2(a+b)=\ell^2+m^2$}$$