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 $C{P}^2+C{Q}^2+C{R}^2+C{S}^2=$

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

Answer 1

Answer: (D)

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
$x^2+\frac{1}{x^2}+2gx+\frac{2fy}{x}+c=0$
$\Rightarrow x^4+2g{x}^3+c{x}^2+2fx+1=0$
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
$y^4+2f{y}^3+c{y}^2+2gy+1=0$
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
$C{P}^2+C{Q}^2+C{R}^2+C{S}^2$
$=x_1^2+x_2^2+x_3^2+x_4^2+y_1^2+y_2^2+y_3^2+y_4^2$
$\left(\text{where}{x}_1{y}_1=1\text{etc}.\right)$
$=4g^2-2c+4f^2-2c$
$4\left(g^2+f^2-c\right)=4r^2$