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If the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{a}^{2}$ intersects the hyperbola $\mathrm{xy}=\mathrm{c}^{2}$ in four points $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right), R\left(x_{3}, y_{3}\right)$ and $S\left(x_{4}, y_{4}\right)$, then
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Verified Answer
The correct answer is:
$\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}+\mathrm{x}_{4}=0$
Given, $x^{2}+y^{2}=a^{2}$ and $x y=c^{2}$
$$
\begin{aligned}
& \therefore & x^{2}+\left(\frac{c^{2}}{x}\right)^{2} &=a^{2} \\
\Rightarrow & & x^{4}-a^{2} x^{2}+c^{4} &=0 \\
& \therefore & x_{1}+x_{2}+x_{3}+x_{4} &=0
\end{aligned}
$$
$$
\begin{aligned}
& \therefore & x^{2}+\left(\frac{c^{2}}{x}\right)^{2} &=a^{2} \\
\Rightarrow & & x^{4}-a^{2} x^{2}+c^{4} &=0 \\
& \therefore & x_{1}+x_{2}+x_{3}+x_{4} &=0
\end{aligned}
$$
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