Q64 / 155 · WBJEE 2017 · Mathematics
The line segment joining the foci of the hyperbola $x^{2}-y^{2}+1=0$ is one of the diameters of a circle. The equation of the circle is
- A$x^{2}+y^{2}=4$
- B$x^{2}+y^{2}=\sqrt{2}$
- Correct answerC$x^{2}+y^{2}=2$
- D$x^{2}+y^{2}=2 \sqrt{2}$
Solution
Given, equation of hyperbola is
$$
x^{2}-y^{2}+1=0
$$
$\Rightarrow \quad y^{2}-x^{2}=1$
On comparing it with $\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1,$ we get
$$
a=b=1
$$
Now, $e=\sqrt{1+\frac{a^{2}}{b^{2}}}=\sqrt{1+\frac{1}{1}}=\sqrt{2}$
$\therefore$ Foci $=(0, \pm b e)=(0, \pm \sqrt{2})$
since, line joining foci of hyperbola is diameter of circle.
$\therefore$ Centre of circle $=\left(\frac{0+0}{2}, \frac{\sqrt{2}-\sqrt{2}}{2}\right)=(0,0)$
$\begin{aligned} \text { and radius } &=\frac{1}{2} \sqrt{(0-0)^{2}+(\sqrt{2}-(-\sqrt{2}))^{2}} \\ &=\frac{1}{2} \sqrt{(2 \sqrt{2})^{2}}=\sqrt{2} \end{aligned}$
$\therefore$ Equation of circle will be
$$
(x-0)^{2}+(y-0)^{2}=(\sqrt{2})^{2}
$$
$\Rightarrow \quad x^{2}+y^{2}=2$
Hence, the correct answer is option: 'C'.