KCET · Maths · Circle
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
- A \(\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}+\mathrm{y}_{4}=2\)
- B \(\mathrm{x}_{1} \mathrm{x}_{2} \mathrm{x}_{3} \mathrm{x}_{4}=2 \mathrm{c}^{4}\)
- C \(\mathrm{y}_{1} \mathrm{y}_{2} \mathrm{y}_{3} \mathrm{y}_{4}=2 \mathrm{c}^{4}\)
- D \(\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}+\mathrm{x}_{4}=0\)
Answer & Solution
Correct Answer
(D) \(\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}+\mathrm{x}_{4}=0\)
Step-by-step Solution
Detailed explanation
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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