Question

For hyperbola $\frac{x^2}{16}-\frac{y^2}{12}=0$ and circle $x^2+y^2=25$, find concyclic points.

• A. $x=\frac{\pm 7}{\sqrt{1}0},y=\frac{\pm 5\sqrt{3}}{\sqrt{7}}$
• B. $x=\frac{\pm 10}{\sqrt{7}},y=\frac{\pm 3\sqrt{5}}{\sqrt{7}}$
• C. $x=\frac{\pm 10}{\sqrt{7}},y=\frac{\pm 5\sqrt{3}}{\sqrt{7}}$
• D. $x=\frac{\pm 5}{\sqrt{7}},y=\frac{\pm 10\sqrt{3}}{\sqrt{7}}$

Answer 1

Answer: (C)

$x=\frac{\pm 10}{\sqrt{7}},y=\frac{\pm 5\sqrt{3}}{\sqrt{7}}$

The equation of hyperbola is $\frac{x^2}{16}-\frac{y^2}{12}=0$

The equation of circle is $x^2+y^2=25$

Now substitute $x^2=\frac{4}{3}{y}^2$ in the circle equation

We get $\frac{7}{3}{y}^2=25$

$\Rightarrow y=\pm \frac{5\sqrt{3}}{\sqrt{7}}$

Therefore $x=\pm \frac{10}{\sqrt{7}}$

Therefore the correct option is $C$