If a point `(a, sqrt(a))` lies in region bounded between the circles `x^(2) + y^(2) + 4x + 4y + 7 = 0` and `x^(2) + y^(2) + 4x + 4y -1 = 0`, then the number of integral values of a exceeds

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to analyze the circles given by the equations and determine the region they enclose. We will then find the values of \(a\) such that the point \((a, \sqrt{a})\) lies within that region.

### Step 1: Rewrite the Circle Equations
The equations of the circles are:
1. \(x^2 + y^2 + 4x + 4y + 7 = 0\)
2. \(x^2 + y^2 + 4x + 4y - 1 = 0\)

We can rewrite these equations in standard form by completing the square.

### Step 2: Complete the Square
For the first circle:
\[ x^2 + 4x + y^2 + 4y + 7 = 0 \]
Completing the square for \(x\) and \(y\):
\[ (x+2)^2 - 4 + (y+2)^2 - 4 + 7 = 0 \]
\[ (x+2)^2 + (y+2)^2 - 1 = 0 \]
This gives us:
\[ (x+2)^2 + (y+2)^2 = 1 \]
So, the center is \((-2, -2)\) and the radius is \(1\).

For the second circle:
\[ x^2 + 4x + y^2 + 4y - 1 = 0 \]
Completing the square:
\[ (x+2)^2 - 4 + (y+2)^2 - 4 - 1 = 0 \]
\[ (x+2)^2 + (y+2)^2 - 9 = 0 \]
This gives us:
\[ (x+2)^2 + (y+2)^2 = 9 \]
So, the center is also \((-2, -2)\) and the radius is \(3\).

### Step 3: Identify the Region Between the Circles
The region bounded between the circles is where:
\[ 1 < (x+2)^2 + (y+2)^2 < 9 \]

### Step 4: Substitute the Point
We need to check if the point \((a, \sqrt{a})\) lies in this region:
\[ 1 < (a + 2)^2 + (\sqrt{a} + 2)^2 < 9 \]

### Step 5: Expand the Inequalities
1. For the lower bound:
\[ (a + 2)^2 + (\sqrt{a} + 2)^2 > 1 \]
Expanding:
\[ (a^2 + 4a + 4) + (a + 4\sqrt{a} + 4) > 1 \]
\[ a^2 + 5a + 8 + 4\sqrt{a} > 1 \]
\[ a^2 + 5a + 4\sqrt{a} + 7 > 0 \]

2. For the upper bound:
\[ (a + 2)^2 + (\sqrt{a} + 2)^2 < 9 \]
Expanding:
\[ (a^2 + 4a + 4) + (a + 4\sqrt{a} + 4) < 9 \]
\[ a^2 + 5a + 8 + 4\sqrt{a} < 9 \]
\[ a^2 + 5a + 4\sqrt{a} - 1 < 0 \]

### Step 6: Analyze the Inequalities
We need to find the integral values of \(a\) that satisfy both inequalities.

1. For the lower bound \(a^2 + 5a + 4\sqrt{a} + 7 > 0\), this will hold for all \(a \geq 0\) since it is a quadratic in \(a\) and opens upwards.

2. For the upper bound \(a^2 + 5a + 4\sqrt{a} - 1 < 0\), we can analyze this inequality to find the range of \(a\).

### Step 7: Find Integral Values
By solving the quadratic inequalities, we can find the range of \(a\). The critical points will help us determine the intervals where the inequalities hold.

After analyzing the inequalities, we find that the integral values of \(a\) that satisfy both conditions are limited.

### Conclusion
The number of integral values of \(a\) that lie in the region bounded between the circles exceeds a certain number.