Question

A rectangular hyperbola whose center is $C$ is cut by any circle of radius $r$ in four points $P,Q,R$ and $S$. Then if,

$C{P}^2+C{Q}^2+C{R}^2+C{S}^2=k{r}^2$, find the value of $k$.

Answer 1

Let the rectangular hyperbola be $xy=c^2$, and the circle be $x^2+y^2=r^2$ with a common center $C(0,0)$

From equation of hyperbola $y=\frac{c^2}{x}$

To find the common points of hyperbola and circle let's put the value of $y$ from equation of hyperbola into equation of circle, we get,

$\Rightarrow x^2+(\frac{c^2}{x}{)}^2=r^2$

$\Rightarrow x^2+\frac{c^4}{x^2}=r^2$

$\Rightarrow (x^2{)}^2-(r^2)x^2+c^4=0$ .... $\left(1\right)$

Taking Eq $\left(1\right)$ as a quadratic equation in $x^2$, the sum of roots of the equation is:

$=(x_1{)}^2+(x_2{)}^2=r^2$ or $(x_3{)}^2+(x_4{)}^2-=r^2$ .... $\left(2\right)$

Similarly if we put the value of $x$ from equation of hyperbola into equation of circle, we get

$\Rightarrow y^2+(\frac{c^2}{y}{)}^2=r^2$

$\Rightarrow y^2+\frac{c^4}{y^2}=r^2$

$\Rightarrow (y^2{)}^2-(r^2)y^2+c^4=0$ .... $\left(3\right)$

Taking Eq $\left(3\right)$ as a quadratic equation in $y^2$, the sum of roots of the equation is:

$(y_1{)}^2+(y_2{)}^2=r^2$ or $(y_3{)}^2+(y_4{)}^2=r^2$ .... $\left(4\right)$

If the center is $C(0,0)$ and $P,Q,R,S$ be $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ and $(x_4,y_4)$ respectively.

Hence $C{P}^2=x_1^2+y_1^2$

$C{Q}^2=x_2^2+y_2^2$

$C{R}^2=x_3^2+y_3^2$

$C{S}^2=x_4^2+y_4^2$

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$

$\Rightarrow 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$ ....$\left(5\right)$

Putting values from eq$\left(2\right)$ and eq.$\left(4\right)$ into eq$\left(5\right)$, we get,

$\Rightarrow C{P}^2+C{Q}^2+C{R}^2+C{S}^2=$$r^2+r^2+r^2+r^2=4r^2$

Hence comparing with $k{r}^2$, $k=4$