Since the equations of tangents x - y - 2 = 0 and x — y + 2 = 0 are parallel.
$\therefore $ Distance between them = diameter of the circle
$= \frac{2-\left(-2\right)}{\sqrt{1^{2} +1^{2}}} \left(\because \frac{C_{2} -C_{1}}{\sqrt{a^{2} +b^{2}}} \right) $
$= \frac{4}{\sqrt{2}} = 2 \sqrt{2} $
$\therefore $ Radius$= \frac{1}{2} \left(2\sqrt{2}\right) = \sqrt{2}$
It is clear from the figure that centre lies on
the origin.
$\therefore $ Equation of circle is
$ \left(x-0\right)^{2} + \left(y-0\right)^{2} = \left(\sqrt{2}\right)^{2} $
$\Rightarrow x^{2} + y^{2 } = 2 $
