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 values of \( a \) such that the point \( (-2a, a + 1) \) lies in the interior of the region bounded by the circle \( x^2 + y^2 = 4 \) and the parabola \( y^2 = 4x \).

### Step 1: Analyze the Circle
The equation of the circle is given by:
\[ x^2 + y^2 = 4 \]
To find the condition for the point \( (-2a, a + 1) \) to lie inside the circle, we substitute \( x = -2a \) and \( y = a + 1 \) into the circle's equation:
\[ (-2a)^2 + (a + 1)^2 < 4 \]
This simplifies to:
\[ 4a^2 + (a^2 + 2a + 1) < 4 \]
Combining the terms gives:
\[ 5a^2 + 2a + 1 < 4 \]
Subtracting 4 from both sides:
\[ 5a^2 + 2a - 3 < 0 \]

### Step 2: Factor the Quadratic
Next, we factor the quadratic inequality:
\[ 5a^2 + 2a - 3 = 0 \]
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 5 \cdot (-3)}}{2 \cdot 5} \]
Calculating the discriminant:
\[ = \frac{-2 \pm \sqrt{4 + 60}}{10} = \frac{-2 \pm \sqrt{64}}{10} = \frac{-2 \pm 8}{10} \]
This gives us the roots:
\[ a = \frac{6}{10} = \frac{3}{5} \quad \text{and} \quad a = \frac{-10}{10} = -1 \]
Thus, we have:
\[ 5a^2 + 2a - 3 < 0 \implies (a + 1)(5a - 3) < 0 \]

### Step 3: Determine the Intervals
To find the intervals where this product is negative, we analyze the sign changes:
- The critical points are \( a = -1 \) and \( a = \frac{3}{5} \).
- Testing intervals:
- For \( a < -1 \): both factors are negative, product is positive.
- For \( -1 < a < \frac{3}{5} \): first factor is positive, second is negative, product is negative.
- For \( a > \frac{3}{5} \): both factors are positive, product is positive.

Thus, the solution for the first condition is:
\[ -1 < a < \frac{3}{5} \]

### Step 4: Analyze the Parabola
The equation of the parabola is:
\[ y^2 = 4x \]
To find the condition for the point \( (-2a, a + 1) \) to lie inside the parabola, we substitute:
\[ (a + 1)^2 < 4(-2a) \]
This simplifies to:
\[ a^2 + 2a + 1 < -8a \]
Rearranging gives:
\[ a^2 + 10a + 1 < 0 \]

### Step 5: Factor the Quadratic
Now we solve the quadratic inequality:
\[ a^2 + 10a + 1 = 0 \]
Using the quadratic formula:
\[ a = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-10 \pm \sqrt{100 - 4}}{2} = \frac{-10 \pm \sqrt{96}}{2} = \frac{-10 \pm 4\sqrt{6}}{2} \]
This simplifies to:
\[ a = -5 \pm 2\sqrt{6} \]
Thus, the roots are:
\[ a_1 = -5 - 2\sqrt{6}, \quad a_2 = -5 + 2\sqrt{6} \]

### Step 6: Determine the Intervals
The solution for the second condition is:
\[ -5 - 2\sqrt{6} < a < -5 + 2\sqrt{6} \]

### Step 7: Combine the Intervals
We need to find the intersection of the two intervals:
1. From the circle: \( -1 < a < \frac{3}{5} \)
2. From the parabola: \( -5 - 2\sqrt{6} < a < -5 + 2\sqrt{6} \)

Calculating \( -5 + 2\sqrt{6} \):
- Since \( \sqrt{6} \approx 2.45 \), we have \( -5 + 4.9 \approx -0.1 \).

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

### Final Answer
The final solution is:
\[ -1 < a < -5 + 2\sqrt{6} \]