Consider a curve $ax^2 + 2hxy + by^2 = 1$ and a point P not on the curve. A line is drawn from the point P which intersects the curve at points $Q$ and $R$. If the product $PQ\cdot PR$ is independent of the slope of the line, then show that the curve is a circle.
My Attempt I know $PQ\cdot PR = PT^2$ for a circle where T is point of contact of tangent to circle. But I cannot prove the reverse of it.
Hint:
Let the point P be $(h,k)$
Since Q and R refer to signed distances:
Let Q be $(h+r_1$ $cos\theta$, $h+r_1$ $sin\theta)$ and R be $(h+r_2$ $cos\theta$, $h+r_2$ $sin\theta)$
Note that Q and R both lie on the circle, and hence if we replace $(x,y)$ in the equation of circle with $(h+r$ $cos\theta$, $h+r$ $sin\theta)$, we will get a quadratic in $r$, the roots of which will be $r_1$ and $r_2$
We also know that $PQ.PR=PT^2$, where $PQ$ is $r_1$, $PR$ is $r_2$
Substitute and find the quadratic in $r$, the product of roots $(=\frac{c}{a})$ of which is $PQ.PR$. Since the product of the roots is given independent of $\theta$, make the denominator independent of $\theta$. In doing so, you'll see that $a=b$ and $h=0$, which will prove that the required locus is that of a circle.