Question

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. $x^{2}+y^{2}=\sqrt{2}$
• C. $x^2 + y^2 = 2$
• D. $x^{2}+y^{2}=2\sqrt{2}$

Answer 1

Answer: (C)

$x^2 + y^2 = 2$

Answer 2

Given, equation of hyperbola is
$x^{2}-y^{2}+1 =0$
$\Rightarrow y^{2}-x^{2} =1$
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 $=\left(\frac{0+0}{2}, \frac{\sqrt{2}-\sqrt{2}}{2}\right)=(0,0)$
and radius $=\frac{1}{2} \sqrt{(0-0)^{2}+(\sqrt{2}-(-\sqrt{2}))^{2}}$
$=\frac{1}{2} \sqrt{(2 \sqrt{2})^{2}}=\sqrt{2}$
$\therefore$ Equation of circle will be
$(x-0)^{2}+(y-0)^{2} =(\sqrt{2})^{2}$
$\Rightarrow x^{2}+y^{2} =2$