Question

For hyperbola $\frac{x^2}{144}-\frac{y^2}{100}=1$, circle $x^2+y^2=169$, concyclic points are

• A. $\pm \frac{\sqrt{269}}{\sqrt{61}},\pm \frac{5}{\sqrt{61}}$
• B. $\pm \frac{6\sqrt{29}}{\sqrt{61}},\pm \frac{25}{\sqrt{61}}$
• C. $\pm \frac{6\sqrt{269}}{\sqrt{61}},\pm \frac{25}{\sqrt{61}}$
• D. None

Answer 1

The correct option is C $\pm \frac{6\sqrt{269}}{\sqrt{61}},\pm \frac{25}{\sqrt{61}}$

The given hyperbola $\frac{x^2}{144}-\frac{y^2}{100}=1$ and given circle $x^2+y^2=169$, cut each-other at four common points, which are con-cyclic points.

From equation of circle getting $y^2=169-x^2$ ....$\left(1\right)$

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

$\to \frac{x^2}{144}-\frac{(169-x^2)}{100}=1$

$\to \frac{x^2}{144}+\frac{x^2}{100}-\frac{\left(169\right)}{100}=1$

$\to \frac{244x^2}{14400}=\frac{\left(269\right)}{100}$

$\to x=\pm \frac{12\sqrt{269}}{2\sqrt{61}}$

$\to x=\pm \frac{6\sqrt{269}}{\sqrt{61}}$

Now $y^2=169-x^2=169-$ $(\pm \frac{6\sqrt{269}}{\sqrt{61}}{)}^2$

$\to y^2=\left(\frac{10309-9684}{61}\right)$

$\to y^2=\pm \frac{625}{61}$

$\to y=\pm \frac{25}{\sqrt{61}}$

Hence con-cyclic point of given hyperbola and circle are $\to \pm \frac{6\sqrt{269}}{\sqrt{61}}$, $\pm \frac{25}{\sqrt{61}}$

Correct option is $C$.