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

Options

A. `(-oo, -2-(1)/(sqrt(2)))uu(-2+(1)/(sqrt(2)),oo)`
B. `(-2+(1)/(sqrt(2)),2+(1)/(sqrt(2)))`
C. `(-oo, 1-(1)/(sqrt(2)))uu(1+(1)/(sqrt(2)),oo)`
D. `(1-(1)/(sqrt(2)), 1+(1)/(sqrt(2)))`

Explanation

To solve the problem step by step, we need to analyze the given information and derive the necessary conditions for the line \( y = x + 1 \) to intersect the circles at real points.

### Step 1: Identify the radius of the smallest circle
The equation of the smallest circle is given as:
\[ x^2 + y^2 = 4 \]
From this equation, we can see that the radius \( r_1 \) of the smallest circle is:
\[ r_1 = \sqrt{4} = 2 \]

### Step 2: Define the radius of the larger circle
Let the radius of the larger circle be \( r_2 \). According to the problem, the difference in radii \( d \) is defined as:
\[ d = r_2 - r_1 \]
Thus, we can express the radius of the larger circle as:
\[ r_2 = r_1 + d = 2 + d \]

### Step 3: Write the equation of the larger circle
The equation of the larger circle can be expressed as:
\[ x^2 + y^2 = r_2^2 = (2 + d)^2 \]

### Step 4: Substitute the line equation into the circle equation
We substitute \( y = x + 1 \) into the equation of the larger circle:
\[ x^2 + (x + 1)^2 = (2 + d)^2 \]
Expanding this gives:
\[ x^2 + (x^2 + 2x + 1) = (2 + d)^2 \]
\[ 2x^2 + 2x + 1 = (2 + d)^2 \]

### Step 5: Rearranging the equation
Rearranging the equation, we get:
\[ 2x^2 + 2x + 1 - (2 + d)^2 = 0 \]
Let’s denote \( P = 1 - (2 + d)^2 \):
\[ 2x^2 + 2x + P = 0 \]

### Step 6: Condition for real roots
For the quadratic equation \( 2x^2 + 2x + P = 0 \) to have real roots, the discriminant must be greater than or equal to zero:
\[ b^2 - 4ac \geq 0 \]
Here, \( a = 2 \), \( b = 2 \), and \( c = P \):
\[ (2)^2 - 4(2)(P) \geq 0 \]
\[ 4 - 8P \geq 0 \]
This simplifies to:
\[ P \leq \frac{1}{2} \]

### Step 7: Substitute back for P
Substituting back for \( P \):
\[ 1 - (2 + d)^2 \leq \frac{1}{2} \]
This leads to:
\[ -\frac{1}{2} \leq (2 + d)^2 - 1 \]
Thus:
\[ (2 + d)^2 \geq \frac{1}{2} \]

### Step 8: Solving the inequality
Taking square roots gives us two inequalities:
\[ 2 + d \geq \frac{1}{\sqrt{2}} \quad \text{or} \quad 2 + d \leq -\frac{1}{\sqrt{2}} \]
From the first inequality:
\[ d \geq \frac{1}{\sqrt{2}} - 2 \]
From the second inequality:
\[ d \leq -\frac{1}{\sqrt{2}} - 2 \]

### Conclusion
Thus, the difference in radii \( d \) lies in the interval:
\[ d \in \left(-\infty, -\frac{1}{\sqrt{2}} - 2\right) \cup \left(\frac{1}{\sqrt{2}} - 2, \infty\right) \]