Animated Solution for Mathematics - Conic Sections: 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

Welcome, future engineers! Today, we embark on a journey into the elegant world of coordinate geometry. Imagine standing on a vast, flat plane.

In front of you, two curves are drawn: a perfect circle defined by $x^2+y^2=a^2$, centered at the origin with radius $a$, and a rectangular hyperbola $xy=c^2$, which gracefully curves through the first and third quadrants.

These two shapes meet at four distinct points, which we label $P(x_1,y_1)$, $Q(x_2,y_2)$, $R(x_3,y_3)$, and $S(x_4,y_4)$. Our mission is to uncover the hidden relationships between these coordinates.

To find these intersection points, we must solve the two equations simultaneously. From the hyperbola, we have $y=\frac{c^2}{x}$.

Now, let's substitute this into the circle's equation:

This substitution is the bridge between geometry and algebra. When we expand this, we get $x^2+\frac{c^4}{x^2}=a^2$.

To clear the denominator, we multiply the entire equation by $x^2$, leading us to the following expression:

Rearranging the terms, we arrive at a beautiful quartic equation: $x^4-a^2{x}^2+c^4=0$. This is the heart of the problem.

A fourth-degree polynomial, by the Fundamental Theorem of Algebra, must have four roots. These roots, $x_1,x_2,x_3,x_4$, are precisely the $x$-coordinates of our four intersection points.

Now, we invoke the legendary Vieta's formulas. Let's write our quartic equation in its full form:

The sum of the roots is given by the negative of the coefficient of $x^3$ divided by the coefficient of $x^4$. Since the $x^3$ term is missing, its coefficient is $0$, so $x_1+x_2+x_3+x_4=0$.

Similarly, the product of the roots is the constant term divided by the coefficient of $x^4$, which gives us $x_1{x}_2{x}_3{x}_4=c^4$.

Because the original equations are symmetric with respect to $x$ and $y$, we can apply the same logic to the $y$-coordinates. If we had substituted $x=\frac{c^2}{y}$ into the circle's equation, we would have obtained the identical quartic equation:

Thus, by the same logic, $y_1+y_2+y_3+y_4=0$ and $y_1{y}_2{y}_3{y}_4=c^4$.

We have successfully decoded the geometry of these curves, proving that all the given options are correct. Keep practicing, and you will see that math is not just about numbers; it is about finding the hidden symmetry in the universe.