A common tangent to the circles `x^(2)+y^(2)=4` and `(x-3)^(2)+y^(2)=1`, is

Options

A. x=4
B. y=2
C. `x+sqrt(3)y=4`
D. `x+2sqrt(2)y=6`

Explanation

To find the common tangent to the circles given by the equations \(x^2 + y^2 = 4\) and \((x - 3)^2 + y^2 = 1\), we will follow these steps:

### Step 1: Identify the centers and radii of the circles
1. **Circle 1**: The equation \(x^2 + y^2 = 4\) can be rewritten as:
\[ (x - 0)^2 + (y - 0)^2 = 2^2 \]
- Center \(C_1 = (0, 0)\)
- Radius \(r_1 = 2\)

2. **Circle 2**: The equation \((x - 3)^2 + y^2 = 1\) can be rewritten as:
\[ (x - 3)^2 + (y - 0)^2 = 1^2 \]
- Center \(C_2 = (3, 0)\)
- Radius \(r_2 = 1\)

### Step 2: Write the slope form of the tangent
The slope form of the tangent to a circle with center \((\alpha, \beta)\) and radius \(r\) is given by:
\[ y - \beta = m(x - \alpha) \pm r\sqrt{1 + m^2} \]
- For Circle 1:
\[ y = mx + 2\sqrt{1 + m^2} \]
- For Circle 2:
\[ y = m(x - 3) \pm \sqrt{1 + m^2} \]
Simplifying this gives:
\[ y = mx - 3m \pm \sqrt{1 + m^2} \]

### Step 3: Set the equations equal
Since we are looking for a common tangent, we can equate the two expressions for \(y\):
\[ mx + 2\sqrt{1 + m^2} = mx - 3m + \sqrt{1 + m^2} \]

### Step 4: Solve for \(m\)
Rearranging gives:
\[ 2\sqrt{1 + m^2} - \sqrt{1 + m^2} = -3m \]
This simplifies to:
\[ \sqrt{1 + m^2} = -3m \]
Squaring both sides:
\[ 1 + m^2 = 9m^2 \]
This simplifies to:
\[ 8m^2 = 1 \quad \Rightarrow \quad m^2 = \frac{1}{8} \quad \Rightarrow \quad m = \frac{1}{2\sqrt{2}} \text{ or } m = -\frac{1}{2\sqrt{2}} \]

### Step 5: Substitute \(m\) back into the tangent equation
Using \(m = \frac{1}{2\sqrt{2}}\):
1. For Circle 1:
\[ y = \frac{1}{2\sqrt{2}}x + 2\sqrt{1 + \left(\frac{1}{2\sqrt{2}}\right)^2} \]
\[ = \frac{1}{2\sqrt{2}}x + 2\sqrt{1 + \frac{1}{8}} = \frac{1}{2\sqrt{2}}x + 2\sqrt{\frac{9}{8}} = \frac{1}{2\sqrt{2}}x + \frac{3}{2\sqrt{2}} \]
\[ = \frac{1}{2\sqrt{2}}x + \frac{3}{2\sqrt{2}} = \frac{1}{2\sqrt{2}}(x + 3) \]

2. For Circle 2:
Using \(m = -\frac{1}{2\sqrt{2}}\):
\[ y = -\frac{1}{2\sqrt{2}}x + 2\sqrt{1 + \left(-\frac{1}{2\sqrt{2}}\right)^2} \]
\[ = -\frac{1}{2\sqrt{2}}x + 2\sqrt{\frac{9}{8}} = -\frac{1}{2\sqrt{2}}x + \frac{3}{2\sqrt{2}} \]
\[ = -\frac{1}{2\sqrt{2}}(x - 3) \]

### Final Equation of the Common Tangent
The common tangents are:
1. \(x - 2\sqrt{2}y + 6 = 0\) (for positive slope)
2. \(x + 2\sqrt{2}y - 6 = 0\) (for negative slope)