Question

For the curves $x^2+y^2=1$ and $x^2-y^2=4$, which of the following is/are true $?$

Answer 1

Equation of Hyperbola
$H:\frac{x^2}{4}-\frac{y^2}{4}=1\dots \left(1\right)$
Equation of circle
$C:x^2+y^2=1\dots \left(2\right)$
Equation of tangent to $\left(1\right)$ in slope form
$y=mx\pm \sqrt{4m^2-4}$
$y=mx\pm 2\sqrt{m^2-1}\dots \left(3\right)$



Let $\left(3\right)$ is common tangent
$\Rightarrow \left(3\right)$ is tangent to circle
By condition of tangency
Length of perpendicular from
$(0,0)$ to tangent $=$ radius of circle

$\Rightarrow \frac{m\left(0\right)-0\pm 2\sqrt{m^2-1}}{\sqrt{m^2+1}}=1$

$\Rightarrow \pm 2\sqrt{m^2-1}=\sqrt{m^2+1}$

$\Rightarrow 4(m^2-1)=(m^2+1)$

$\Rightarrow m=\pm \sqrt{\frac{5}{3}}$
Equation of tangent
For $m=\sqrt{\frac{5}{3}}$
$y=\sqrt{\frac{5}{3}}x\pm 2\sqrt{\frac{2}{3}}$

$\Rightarrow \sqrt{3}y=\sqrt{5}x\pm 2\sqrt{2}\dots \left(4\right)$

For $m=-\sqrt{\frac{5}{3}}$

$y=-\sqrt{\frac{5}{3}}x\pm 2\sqrt{\frac{2}{3}}$

$\Rightarrow \sqrt{3}y=-\sqrt{5}x\pm 2\sqrt{2}\dots \left(5\right)$

By $\left(4\right)\&\left(5\right)$, there are $4$ common tangents
Option $\left(A\right)\&\left(C\right)$ are correct
There are $4$ common tangents
$\sqrt{5}x-\sqrt{3}y=2\sqrt{2},\sqrt{5}x-\sqrt{3}y=-2\sqrt{2},\sqrt{5}x+\sqrt{3}y=2\sqrt{2},\sqrt{5}x+\sqrt{3}y=-2\sqrt{2}$
Area of region enclosed by these common tangents
$=4\times (\frac{1}{2}\times 2\sqrt{\frac{2}{3}}\times 2\sqrt{\frac{2}{5}})$
$=\frac{16}{\sqrt{15}}$
Also, the common tangents do not touches Hyperbola at $(\sqrt{\frac{5}{2}},\sqrt{\frac{3}{2}})$ as this point does not satisfy the equation of hyperbola.
i.e. ${\left(\sqrt{\frac{5}{2}}\right)}^2-{\left(\sqrt{\frac{3}{2}}\right)}^2=1\ne 4$