If the line `y=2x+c` is a tangent to the circle `x^(2)+y^(2)=5` then a value of c is

Options

A. 3
B. 2
C. 5
D. 4

Answer

To find the value of \( c \) such that the line \( y = 2x + c \) is a tangent to the circle given by the equation \( x^2 + y^2 = 5 \), we can follow these steps:

### Step 1: Identify the parameters
The equation of the line is given as:
\[ y = 2x + c \]
From this equation, we can identify the slope \( m \) of the line as \( 2 \).

The equation of the circle is:
\[ x^2 + y^2 = 5 \]
This can be compared to the standard form \( x^2 + y^2 = r^2 \), where \( r^2 = 5 \). Thus, the radius \( r \) of the circle is:
\[ r = \sqrt{5} \]

### Step 2: Use the tangent condition
For a line to be tangent to a circle, the distance \( d \) from the center of the circle to the line must be equal to the radius of the circle. The center of the circle \( (0, 0) \) and the distance \( d \) from the point \( (0, 0) \) to the line \( Ax + By + C = 0 \) can be calculated using the formula:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For our line \( y = 2x + c \), we can rewrite it in the standard form:
\[ 2x - y + c = 0 \]
Here, \( A = 2 \), \( B = -1 \), and \( C = c \).

### Step 3: Calculate the distance
Substituting \( (x_1, y_1) = (0, 0) \) into the distance formula gives:
\[ d = \frac{|2(0) - 1(0) + c|}{\sqrt{2^2 + (-1)^2}} = \frac{|c|}{\sqrt{4 + 1}} = \frac{|c|}{\sqrt{5}} \]

### Step 4: Set the distance equal to the radius
Since the line is tangent to the circle, we set the distance equal to the radius:
\[ \frac{|c|}{\sqrt{5}} = \sqrt{5} \]

### Step 5: Solve for \( c \)
To eliminate the fraction, multiply both sides by \( \sqrt{5} \):
\[ |c| = 5 \]
This gives us two possible values for \( c \):
\[ c = 5 \quad \text{or} \quad c = -5 \]

### Conclusion
Thus, the values of \( c \) such that the line is tangent to the circle are \( c = 5 \) and \( c = -5 \).