Question

For hyperbola $xy=81$ ,and circle $x^2+y^2=162$, the concyclic points are

• A. $(\pm 81,\pm 81)$
• B. $(\pm 3,3)$
• C. $(\pm 5,\pm 2)$
• D. $(\pm 9,\pm 9)$

Answer 1

Answer: (D)

$(\pm 9,\pm 9)$

Given hyperbola is $xy=81$ ..... $\left(1\right)$

and given circle is $x^2+y^2=162$ ......$\left(2\right)$

The given circle and the hyperbola cut each-other at four points, the common points of circle and hyperbola are called con-cyclic points.

Squaring both sides of equation $\left(1\right)$ we get, $x^2{y}^2=(81{)}^2$ ...$\left(3\right)$

Now putting value of $y^2$ from eq. $\left(2\right)$ into eq. $\left(3\right)$ we get,

$\to x^2(162-x^2)=(81{)}^2$

$\Rightarrow$ $(x^2{)}^2-162x^2+(81{)}^2=0$

$\Rightarrow$ $(x^2{)}^2-2x^2\times 81+(81{)}^2=0$

$\Rightarrow$ $(x^2-81{)}^2=0$

$\Rightarrow$ $x^2=81$

$\Rightarrow x=\pm 9$

Putting the values of $x$ in eq. $\left(1\right)$ we get, $y=\frac{81}{\pm 9}$

$\Rightarrow y=\pm 9$

Hence the four common points or con-cyclic points are $(\pm 9,\pm 9)$

So the correct option is $D$.