Question

A rectangular hyperbola whose centre is C is cut by any circle of radius r in four points P, Q, R and S. Then $CP^2 + CQ^2 + CR^2 + CS^2 =$

• A. $r^2$
• B. $2r^2$
• C. $3r^2$
• D. $4r^2$

Answer 1

Answer: (D)

$4r^2$

Answer 2

Let the equation of the circle be $x^{2}+y^{2}+2gx+2fy+c = 0$ and let the rect. hyperbola be $xy = 1$ $\Rightarrow y =\frac{1}{x}$ Putting in the equation of the circle $x^{2}+\frac{1}{x^{2}}+2gx+\frac{2fy}{x}+c = 0$ $\Rightarrow x^{4}+2gx^{3}+cx^{2}+2fx+1=0$ this is fourth degree equation in $x$ giving four values of $x$ say $x_{1}, x_{2}, x_{3}, x_{4}$ $\therefore\Sigma x_{1} = -2g$ $\Sigma x_{1}x_{2} = c$ $\therefore \Sigma x_{1}^{2} = \left(\Sigma x_{1}\right)^{2} -2 \Sigma x_{1}x_{2} = 4g^{2} -2c$ Equation whose roots are reciprocals of the roots of $\left(1\right)$is $y^{4}+2fy^{3}+cy^{2}+2gy+1 = 0$ If $y_{1}, y_{2}, y_{3}, y_{4}$ are it's roots then $\Sigma y_{1}^{2} = \left( \Sigma y_{1}\right)^{2} -2 \Sigma y_{1}y_{2} = 4f^{2} - 2c$ Now $CP^{2} +CQ^{2}+CR^{2}+CS^{2}$ $= x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+y_{1}^{2}+y_{2}^{2} +y_{3}^{2}+y_{4}^{2}$ $\left( {\text{where}}\,\, x_{1}y_{1} = 1 \,\,{\text{etc}}. \right)$ $= 4g^{2}-2c+4f^{2}-2c$ $4\left(g^{2}+f^{2}-c\right) = 4r^{2}$