Question

For hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=1$, and circle $x^2+y^2=100$, con-cyclic points are

• A. $\pm \frac{10\sqrt{29}}{\sqrt{4}0},\pm \frac{20\sqrt{3}}{\sqrt{40}}$
• B. $\pm \frac{10\sqrt{29}}{\sqrt{41}},\pm \frac{20\sqrt{3}}{\sqrt{41}}$
• C. $\pm \frac{10}{\sqrt{4}1},\pm \frac{20\sqrt{3}}{\sqrt{41}}$
• D. $\pm \frac{\sqrt{29}}{\sqrt{4}1},\pm \frac{10\sqrt{3}}{\sqrt{41}}$

Answer 1

The correct option is B $\pm \frac{10\sqrt{29}}{\sqrt{41}},\pm \frac{20\sqrt{3}}{\sqrt{41}}$

The given hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=1$ and given circle $x^2+y^2=100$, cut each-other at four common points, which are con-cyclic points.

From equation of circle getting $y^2=100-x^2$ ....$\left(1\right)$

and putting value of $y^2$ from equation $\left(1\right)$ into equation of given hyperbola we get,

$\to \frac{x^2}{25}-\frac{(100-x^2)}{16}=1$

$\to \frac{x^2}{25}+\frac{x^2}{16}-\frac{\left(100\right)}{16}=1$

$\to \frac{41x^2}{25\times 16}=\frac{\left(29\right)}{4}$

$\to x=\pm \frac{10\sqrt{29}}{\sqrt{41}}$

Hence,

$x_1=+\frac{10\sqrt{29}}{\sqrt{41}}$

$x_2=-\frac{10\sqrt{29}}{\sqrt{41}}$

Now $y^2=100-x^2=100-$ $(\pm \frac{10\sqrt{29}}{\sqrt{41}}{)}^2$

$\to y^2=100\left(\frac{41-29}{41}\right)$

$\to y=\pm \frac{20\sqrt{3}}{41}$

So $y_1=+\frac{20\sqrt{3}}{41}$

$y_2=-\frac{20\sqrt{3}}{41}$

Hence con-cyclic point of given hyperbola and circle are:

$\to \pm \frac{10\sqrt{29}}{\sqrt{41}}$, $\pm \frac{20\sqrt{3}}{41}$

Correct option is $B$.