If the circle $x^2+y^2=a^2$ intersects the hyperbola $x y=b^2$ at four points $\left(x_1, y_1\right)$, $\left(x_2, y_2\right),\left(x_3, y_3\right),\left(x_4, y_4\right)$, then $y_1 y_2 y_3 y_4=$

• A

  $a^4$
• B

  0
• Correct answer
  C

  $b^4$
• D

  $b^2$

Solution

We have, $x^2+y^2=a^2$ $\ldots(\mathrm{i})$ and $x y=b^2$ $\ldots(\mathrm{ii})$
Let four points $A\left(x_1, y_1\right), B\left(x_2, y_2\right), C\left(x_3, y_3\right)$ and $D\left(x_4, y_4\right)$ $\therefore \quad O A^2=x_1^2+y_1^2$ $O B^2=x_2^2+y_2^2$ $O C^2=x_3^2+y_3^2$ and $\quad O D^2=x_4^2+y_4^2$ Such that $O A^2+O B^2+O C^2+O D^2$ $=\Sigma x_1^2+\Sigma y_1^2=4 a^2$ From Eq. (ii) $x=\frac{b^2}{y}$ From Eq. (i) $\left(\frac{b^2}{y}\right)^2+y^2=a^2$ $\Rightarrow \quad b^4+y^4=a^2 y^2$ $\Rightarrow \quad y^4-a^2 y^2+b^4=0$ This is an equation of 4 th degree in $y$ and its four roots are $y_1, y_2, y_3, y_4$, then $y_1 y_2 y_3 y_4=b^4$

Hence, the correct answer is option: 'C'.