<s>[INST] If a variable chord of the hyperbola subtend a right angle at the centre, find the radius of the circle it is tangent to 
If a variable line $x\cos\alpha+y\sin\alpha=p$ which is a chord of the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, $(b>a)$ subtend a right angle at the centre of hyperbola,then prove that it always touches a fixed circle whose radius is $\dfrac{ab}{\sqrt{b^2-a^2}}$


$x\cos\alpha+y\sin\alpha=p$ touches(tangent to) a fixed circle $x^2+y^2=p^2$ of radius $p$. So I need to find $p$.
Let $A$,$B$ be the points where the given lines intersects the hyperbola, ie. $\angle AOB=90^\circ$
How do I proceed further and find the radius of the required circle ?
Note: I really do not want to use some existing shortcut formula here
 [/INST] Let $A=(x_1,y_1)$ and $B=(x_2,y_2)$ be the intersections of line $\cos+\sin=$ with the hyperbola. We have $\angle AOB=90°$ if $x_1x_2+y_1y_2=0$. Substituting there 
$$y={p-x\cos\alpha\over\sin\alpha}$$ 
we thus obtain:
$$
\tag{*}
x_1x_2-(x_1+x_2)p\cos\alpha+p^2=0.
$$
But $x_1$ and $x_2$ are the solutions of the equation
$$
{x^2\over a^2}-{(p-x\cos\alpha)^2/\sin^2\alpha\over b^2}=1
$$
which can be rearranged to
$$
(b^2\sin^2\alpha-a^2\cos^2\alpha)x^2+2a^2p\cos\alpha\ x-a^2(p^2+b^2\sin^2\alpha)=0.
$$
From this equation we get:
$$
x_1+x_2=-a^2{2p\cos\alpha\over b^2\sin^2\alpha-a^2\cos^2\alpha},
\quad
x_1x_2=-a^2{p^2+b^2\sin^2\alpha\over b^2\sin^2\alpha-a^2\cos^2\alpha}.
$$
Substituting these into (*) dependence on $\alpha$ cancels out and we can solve for $p$.
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