Two concentric circles of which smallest is $x^{2} + y^{2} = 4$ , have the difference in radii as d, if line $y = x + 1$ cuts the circles in real points, then d lies in the interval

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$(- \infty, 1 - \frac{1}{\sqrt{2}}) \cup (1 + \frac{1}{\sqrt{2}}, \infty)$

$(1 - \frac{1}{\sqrt{2}}, 1 + \frac{1}{\sqrt{2}})$

$(- 2 + \frac{1}{\sqrt{2}}, 2 + \frac{1}{\sqrt{2}})$

$(- \infty, - 2 - \frac{1}{\sqrt{2}}) \cup (- 2 + \frac{1}{\sqrt{2}}, \infty)$

answer is A.

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Detailed Solution

It is given that the difference in the radii of two concentric circles is d. So, let the equations be

$x^{2} + y^{2} = 4$

and $x^{2} + y^{2} = (2 + d)^{2}$

If the line $y = x + l$ cuts the circles in real points, then

$x^{2} + (x + 1)^{2} = (2 + d)^{2}$ must have real roots.

$\begin{array}{l} ∴ 4 - 8 + 8 (2 + d)^{2} > 0 \\ \Rightarrow 2 (2 + d)^{2} - 1 > 0 \\ \Rightarrow (2 + d)^{2} - {(\frac{1}{\sqrt{2}})}^{2} > 0 \\ \Rightarrow (2 + d - \frac{1}{\sqrt{2}}) (2 + d + \frac{1}{\sqrt{2}}) > 0 \\ \Rightarrow d < - 2 - \frac{1}{\sqrt{2}} or, d > - 2 + \frac{1}{\sqrt{2}} \\ \Rightarrow d \in (- \infty, - 2 - \frac{1}{\sqrt{2}}) \cup (- 2 + \frac{1}{\sqrt{2}}, \infty) \end{array}$