The line `y = mx + sqrt( 4 + 4m^(2)), m in R ` , is a tangent to the circle

Options

A. `x^(2) + y^(2) = 2`
B. `x^(2)+y^(2) = 4`
C. `x^(2)+ y^(2)= 1`
D. None of these

Answer

To determine the equation of the circle for which the line \( y = mx + \sqrt{4 + 4m^2} \) is a tangent, we can follow these steps:

### Step 1: Identify the form of the line
The line is given as:
\[ y = mx + \sqrt{4 + 4m^2} \]
Here, we can identify \( c = \sqrt{4 + 4m^2} \).

### Step 2: Use the condition for tangency
For a line \( y = mx + c \) to be a tangent to the circle \( x^2 + y^2 = a^2 \), the condition is:
\[ c = \pm a \sqrt{1 + m^2} \]
In our case, we have:
\[ \sqrt{4 + 4m^2} = \pm 2 \sqrt{1 + m^2} \]

### Step 3: Square both sides to eliminate the square root
Squaring both sides gives:
\[ 4 + 4m^2 = 4(1 + m^2) \]

### Step 4: Simplify the equation
Expanding the right side:
\[ 4 + 4m^2 = 4 + 4m^2 \]
This simplifies to:
\[ 4 + 4m^2 = 4 + 4m^2 \]
This is always true, indicating that the line is indeed a tangent to the circle.

### Step 5: Determine the radius and center of the circle
From the condition, we can see that the radius \( a \) of the circle is:
\[ a = 2 \]
And the center of the circle is at \( (0, 0) \).

### Step 6: Write the equation of the circle
The standard form of the equation of a circle with center at \( (0, 0) \) and radius \( 2 \) is:
\[ x^2 + y^2 = 2^2 \]
Thus, the equation of the circle is:
\[ x^2 + y^2 = 4 \]

### Conclusion
The equation of the circle for which the line \( y = mx + \sqrt{4 + 4m^2} \) is a tangent is:
\[ \boxed{x^2 + y^2 = 4} \]