Question

If the circle $x^2+y^2=16$, cuts the rectangular hyperbola $xy=8$ at four points $A,B,C$ and $D,$ then $\frac{(OA{)}^2+(OB{)}^2+(OC{)}^2+(OD{)}^2}{16}$is equal to (where $O$ be the origin.)

Answer 1

If a circle cuts the hyperbola $xy=c^2$at four points $A\left(t_1\right),B\left(t_2\right),$
$C\left(t_3\right)$ and $D\left(t_4\right)$, then
$\therefore O{A}^2+O{B}^2+O{C}^2+O{D}^2=4r^2$, where $r$ is the radius of the circle and $O$ be the origin.
$\therefore O{A}^2+O{B}^2+O{C}^2+O{D}^2=4\times 16=64$
$\Rightarrow \frac{O{A}^2+O{B}^2+O{C}^2+O{D}^2}{16}=4$