If the line `y = 3x + c` is a tangent to `x^(2) + y^(2) = 4` then the value of c is

Options

A. `pm4`
B. `pm2sqrt(10)`
C. `pm10sqrt(2)`
D. `pmsqrt(10)`

Answer

To find the value of \( c \) such that the line \( y = 3x + c \) is a tangent to the circle given by the equation \( x^2 + y^2 = 4 \), we can follow these steps:

### Step 1: Substitute the line equation into the circle equation
We start by substituting the line equation \( y = 3x + c \) into the circle equation \( x^2 + y^2 = 4 \).

\[ x^2 + (3x + c)^2 = 4 \]

### Step 2: Expand the equation
Next, we expand the equation:

\[ x^2 + (3x + c)(3x + c) = 4 \]

This gives:

\[ x^2 + (9x^2 + 6cx + c^2) = 4 \]

Combining like terms, we have:

\[ 10x^2 + 6cx + (c^2 - 4) = 0 \]

### Step 3: Set the discriminant to zero
For the line to be tangent to the circle, the quadratic equation must have exactly one solution. This occurs when the discriminant is zero. The discriminant \( D \) for the quadratic equation \( ax^2 + bx + c = 0 \) is given by:

\[ D = b^2 - 4ac \]

In our case, \( a = 10 \), \( b = 6c \), and \( c = c^2 - 4 \). Therefore, we set the discriminant equal to zero:

\[ (6c)^2 - 4(10)(c^2 - 4) = 0 \]

### Step 4: Solve the discriminant equation
Expanding the equation gives:

\[ 36c^2 - 40(c^2 - 4) = 0 \]

This simplifies to:

\[ 36c^2 - 40c^2 + 160 = 0 \]

Combining like terms results in:

\[ -4c^2 + 160 = 0 \]

Rearranging gives:

\[ 4c^2 = 160 \]

Dividing both sides by 4 yields:

\[ c^2 = 40 \]

### Step 5: Find the values of \( c \)
Taking the square root of both sides gives:

\[ c = \pm \sqrt{40} \]

This can be simplified to:

\[ c = \pm 2\sqrt{10} \]

### Final Answer:
Thus, the values of \( c \) are:

\[ c = 2\sqrt{10} \quad \text{and} \quad c = -2\sqrt{10} \]

---