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. The first circle \(x^2 + y^2 = 4\) has:
- Center \(C_1(0, 0)\)
- Radius \(r_1 = \sqrt{4} = 2\)

2. The second circle \((x - 3)^2 + y^2 = 1\) has:
- Center \(C_2(3, 0)\)
- Radius \(r_2 = \sqrt{1} = 1\)