Q. If the circle $x^{2} + y^{2} = a^{2}$ intersects the hyperbola $xy = c^{2}$ in four points $P\left( x_{1},\ y_{1} \right),\ Q\left( x_{2},\ y_{2} \right),\ R(x_{3},\ y_{3})$ and $S(x_{4},\ y_{4})$, then
A) $x_{1} + x_{2} + x_{3} + x_{4} = 0$
B) $y_{1} + y_{2} + y_{3} + y_{4} = 0$
C) $x_{1}x_{2}x_{3}x_{4} = c^{4}$
D) All of these
Explanation
(d)
It is given that
$x^{2} + y^{2} = a^{2}$ ..(i)
and $xy = c^{2}$ …(ii)
From Eq.(i) and (ii),
$x^{2} + \frac{c^{4}}{x^{2}} = a^{2}$
$\Rightarrow x^{4} - a^{2}x^{2} + c^{4} = 0$ …(iii)
Now, $x_{1},\ x_{2},\ x_{3},\ x_{4}$ will be roots of Eq.(iii)
$\therefore x_{1} + x_{2} + x_{3} + x_{4} = 0$
and $x_{1}x_{2}x_{3}x_{4} = c^{4}$
Similarly, $y_{1} + y_{2} + y_{3} + y_{4} = 0$
and $y_{1}y_{2}y_{3}y_{4} = c^{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.