If `(-2a,a+1)` lies in the interior (smaller region) bounded by the circle `x^2+y^2=4` and the parabola `y^2=4x`, then

Options

A. `a in (-1,5- 2 sqrt(6))`
B. `a in (1,5+ 2 sqrt(6))`
C. `a in (-1,-5+ 2 sqrt(6))`
D. `a in (-1,-5+4 sqrt(6))`

Answer

To solve the problem, we need to determine the range of the variable \( a \) such that the point \( (-2a, a+1) \) lies in the interior of the circle defined by \( x^2 + y^2 = 4 \) and the parabola defined by \( y^2 = 4x \).

### Step 1: Understand the curves

1. **Circle**: The equation \( x^2 + y^2 = 4 \) represents a circle centered at the origin with a radius of 2.
2. **Parabola**: The equation \( y^2 = 4x \) represents a right-opening parabola.

### Step 2: Determine the conditions for the point to lie within the curves

For the point \( (-2a, a+1) \) to lie in the interior of the circle and the parabola, we need to satisfy the following inequalities:

1. **Inside the Circle**:
\[ x^2 + y^2 < 4 \]
Substituting \( x = -2a \) and \( y = a + 1 \):
\[ (-2a)^2 + (a + 1)^2 < 4 \]
Simplifying:
\[ 4a^2 + (a^2 + 2a + 1) < 4 \]
\[ 5a^2 + 2a + 1 < 4 \]
\[ 5a^2 + 2a - 3 < 0 \]

2. **Inside the Parabola**:
\[ y^2 - 4x < 0 \]
Substituting \( x = -2a \) and \( y = a + 1 \):
\[ (a + 1)^2 - 4(-2a) < 0 \]
Simplifying:
\[ (a^2 + 2a + 1) + 8a < 0 \]
\[ a^2 + 10a + 1 < 0 \]

### Step 3: Solve the inequalities

1. **For the circle**:
We need to solve \( 5a^2 + 2a - 3 < 0 \).
- Finding the roots using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{(2)^2 - 4(5)(-3)}}{2(5)} \]
\[ = \frac{-2 \pm \sqrt{4 + 60}}{10} = \frac{-2 \pm \sqrt{64}}{10} = \frac{-2 \pm 8}{10} \]
This gives us:
\[ a = \frac{6}{10} = \frac{3}{5}, \quad a = \frac{-10}{10} = -1 \]
Thus, the inequality \( 5a^2 + 2a - 3 < 0 \) holds for \( -1 < a < \frac{3}{5} \).

2. **For the parabola**:
We need to solve \( a^2 + 10a + 1 < 0 \).
- Finding the roots:
\[ a = \frac{-10 \pm \sqrt{10^2 - 4(1)(1)}}{2(1)} = \frac{-10 \pm \sqrt{100 - 4}}{2} = \frac{-10 \pm \sqrt{96}}{2} = \frac{-10 \pm 4\sqrt{6}}{2} \]
\[ = -5 \pm 2\sqrt{6} \]
Thus, the roots are \( -5 - 2\sqrt{6} \) and \( -5 + 2\sqrt{6} \). The inequality \( a^2 + 10a + 1 < 0 \) holds between these roots.

### Step 4: Find the intersection of the intervals

Now we need to find the intersection of the two intervals:
1. From the circle: \( -1 < a < \frac{3}{5} \)
2. From the parabola: \( -5 - 2\sqrt{6} < a < -5 + 2\sqrt{6} \)

### Step 5: Determine the final range

To find the intersection, we need to evaluate \( -5 - 2\sqrt{6} \) and \( -5 + 2\sqrt{6} \):
- \( -5 + 2\sqrt{6} \) is approximately \( -5 + 4.89898 \approx -0.10102 \).

Thus, the intersection of the intervals is:
\[ -1 < a < -5 + 2\sqrt{6} \]

### Conclusion

The final range for \( a \) is:
\[ a \in (-1, -5 + 2\sqrt{6}) \]