A) r2
B) 2r2
C) 3r2
D) 4r2
Correct Answer: D
Solution :
Let equation of the rectangular hyperbola be xy =c2 ...(i) and equation of circle be x2 + y2 = r2 ...(ii) Put \[y=\frac{{{c}^{2}}}{x}\]in Eq. (ii), we get \[{{x}^{2}}+\frac{{{c}^{4}}}{{{x}^{2}}}={{r}^{2}}\] \[\Rightarrow \]\[{{x}^{4}}-{{r}^{2}}{{x}^{2}}+{{c}^{4}}=0\]\[\Rightarrow \]\[\sum {{x}_{1}}=0\sum {{x}_{1}}{{x}_{2}}={{r}^{2}}\](iii) Similarly, \[\sum {{y}_{1}}=0,\sum {{y}_{1}}{{y}_{2}}={{\Omega }^{2}}\] ?(iv) Now, \[C{{P}^{2}}+C{{Q}^{2}}+C{{R}^{2}}+C{{S}^{2}}\] \[=x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}+x_{3}^{2}+y_{3}^{2}+x_{4}^{2}+y_{4}^{2}\] \[={{({{x}_{1}}={{x}_{2}}+{{x}_{3}}+{{x}_{4}})}^{2}}-2({{x}_{1}}{{x}_{2}}+{{x}_{2}}{{x}_{3}}+{{x}_{3}}{{x}_{4}}\] \[+{{x}_{4}}{{x}_{1}}+{{x}_{1}}{{x}_{3}}+{{x}_{2}}{{x}_{4}})+{{({{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}})}^{2}}\] \[-2({{y}_{1}}{{y}_{2}}+{{y}_{2}}{{y}_{3}}+{{y}_{3}}{{y}_{4}}+{{y}_{4}}{{y}_{1}}\] \[+{{y}_{1}}{{y}_{3}}+{{y}_{2}}{{y}_{4}})\] \[=0+2{{r}^{2}}+0+2{{r}^{2}}\][from Eqs. (iii) and (iv)] \[=4{{r}^{2}}\]
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