I have an idea...
solve the eqn. of the chord with the eqn. of the circle..
u'll be getting two points of intersection on the circle...
and since the center of the circle is (0,0)
so, u have three vertices fo a right-triangle ... so easy then...!!
The condition that the chord x cos A+y sin A-p=0 of x2+y2-a2=0 may subtend a right angle at the center of the circle is:
(a)a2=2p2
(b)p2=2a2
(c)a=2p
(d)p=2a
I have an idea...
solve the eqn. of the chord with the eqn. of the circle..
u'll be getting two points of intersection on the circle...
and since the center of the circle is (0,0)
so, u have three vertices fo a right-triangle ... so easy then...!!