If the circle x^2 + y^2 = a^2 intersects the | vedclass

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(D) Given the equations $x^2 + y^2 = a^2$ and $xy = c^2$. Substitute $y = \frac{c^2}{x}$ into the circle equation: $x^2 + (\frac{c^2}{x})^2 = a^2$ $x^

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All of the above

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(D) Given the equations $x^2 + y^2 = a^2$ and $xy = c^2$. Substitute $y = \frac{c^2}{x}$ into the circle equation: $x^2 + (\frac{c^2}{x})^2 = a^2$ $x^2 + \frac{c^4}{x^2} = a^2$ $x^4 - a^2x^2 + c^4 = 0$. This is a biquadratic equation in $x$. Comparing it with $x^4 + Bx^3 + Cx^2 + Dx + E = 0$,we have the sum of roots $\sum x_i = 0$ and the product of roots $x_1 x_2 x_3 x_4 = c^4$. Similarly,by substituting $x = \frac{c^2}{y}$,we get $y^4 - a^2y^2 + c^4 = 0$,which gives $\sum y_i = 0$ and $y_1 y_2 y_3 y_4 = c^4$. Thus,all the given options are correct.

If the circle $x^2 + y^2 = a^2$ intersects the hyperbola $xy = c^2$ in four points $P(x_1, y_1), Q(x_2, y_2), R(x_3, y_3), S(x_4, y_4)$,then:

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