Question

If the line $y=mx+c$ is a common tangent to the hyperbola $\frac{x^2}{100}-\frac{y^2}{64}=1$ and the circle $x^2+y^2=36$, then which one of the following is true?

• A. $4c^2=369$
• B. $c^2=369$
• C. $8m+5=0$
• D. $5m=4$

Answer 1

The correct option is A $4c^2=369$

$c=\pm \sqrt{a^2{m}^2-b^2}$
$\Rightarrow c=\pm \sqrt{100m^2-64}\cdots \left(i\right)$

$y=mx\pm \sqrt{100m^2-64}$
$d{|}_{(0,0)}=6$
$\Rightarrow \left|\frac{\sqrt{100m^2-64}}{\sqrt{m^2+1}}\right|=6$
$\Rightarrow 100m^2-64=36m^2+36$
$\Rightarrow 64m^2=100$
$\Rightarrow m=\pm \frac{10}{8}\cdots \left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$
$c^2=100\times \frac{100}{64}-64=\frac{\left(164\right)\left(36\right)}{64}$
$\Rightarrow 4c^2=369$