The line x+2y+a=0 intersects the circle `x^2+y^2-4=0` at two distinct points A and B. Another line 12x-6y-41=0 intersects the circle `x^2+y^2-4x-2y+1=0` at two distincts points C and D.
The value of 'a' so that the line x+2y+a=0 intersects the circle `x^2+y^2-4=0` at two distinct points A and B is

Options

A. `-2sqrt5 lt a lt 2sqrt5`
B. `0 lt a lt 2 sqrt5`
C. `-sqrt5 lt a lt sqrt5`
D. `0 lt a lt sqrt5`

Answer

To find the value of 'a' such that the line \( x + 2y + a = 0 \) intersects the circle \( x^2 + y^2 - 4 = 0 \) at two distinct points, we can follow these steps:

### Step 1: Identify the Circle's Properties
The equation of the circle is given by:
\[ x^2 + y^2 - 4 = 0 \]
This can be rewritten as:
\[ x^2 + y^2 = 4 \]
From this, we can identify the center of the circle as \( (0, 0) \) and the radius \( r = 2 \).

### Step 2: Find the Distance from the Center to the Line
The line is given by:
\[ x + 2y + a = 0 \]
We can rewrite it in the standard form \( Ax + By + C = 0 \) where \( A = 1, B = 2, C = a \).

The distance \( d \) from the center of the circle \( (0, 0) \) to the line can be calculated using the formula:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Substituting the values:
\[ d = \frac{|1(0) + 2(0) + a|}{\sqrt{1^2 + 2^2}} = \frac{|a|}{\sqrt{5}} \]

### Step 3: Set Up the Condition for Two Distinct Intersections
For the line to intersect the circle at two distinct points, the distance \( d \) must be less than the radius \( r \):
\[ \frac{|a|}{\sqrt{5}} < 2 \]

### Step 4: Solve the Inequality
To solve the inequality:
\[ |a| < 2\sqrt{5} \]
This implies:
\[ -2\sqrt{5} < a < 2\sqrt{5} \]

### Step 5: Conclusion
The value of \( a \) must lie within the interval:
\[ a \in (-2\sqrt{5}, 2\sqrt{5}) \]