Question

For hyperbola $\frac{x^2}{25}-\frac{y^2}{12}=1$, and circle $x^2+y^2=100$, Four concyclic points are

• A. $\pm \frac{20}{\sqrt{37}},\pm \frac{30}{\sqrt{3}7}$
• B. $\pm \frac{20\sqrt{7}}{\sqrt{26}},\pm \frac{30}{\sqrt{2}6}$
• C. $\pm \frac{20\sqrt{7}}{\sqrt{37}},\pm \frac{30}{\sqrt{37}}$
• D. $\pm \frac{20\sqrt{7}}{\sqrt{37}},\pm \frac{30\sqrt{7}}{\sqrt{3}7}$

Answer 1

The correct option is C $\pm \frac{20\sqrt{7}}{\sqrt{37}},\pm \frac{30}{\sqrt{37}}$

$\frac{x^2}{25}-\frac{y^2}{12}=1$

$x^2+y^2=100$

$\frac{{100-y}^2}{25}-\frac{y^2}{12}=1$

$4-\frac{y^2}{25}-\frac{y^2}{12}=1$

$43=y^2\times \frac{37}{300}$

$y=\pm \frac{30}{\sqrt{37}}$

$x=\pm \sqrt{100-\frac{900}{37}}=\pm \frac{20\sqrt{7}}{\sqrt{37}}$