Consider the locus of center of the circle which touches the circle `x^(2)+y^(2)=4` externally and the line x=4. The distance of the vertex of the locus from the otigin is __________ .

Explanation

To solve the problem, we need to find the locus of the center of a circle that touches the given circle \(x^2 + y^2 = 4\) externally and the line \(x = 4\). We will then determine the distance of the vertex of this locus from the origin.

### Step-by-Step Solution:

1. **Identify the Given Circle**:
The equation of the given circle is \(x^2 + y^2 = 4\).
- Center: \((0, 0)\)
- Radius: \(r_1 = 2\)

2. **Identify the Given Line**:
The line is given by \(x = 4\).

3. **Let the Center of the Required Circle be \((h, k)\)**:
We need to find the locus of the center of the circle that touches the given circle and the line.

4. **Distance from the Center to the Line**:
The distance from the center \((h, k)\) to the line \(x = 4\) is given by:
\[ d = |h - 4| \]
Since the circle touches the line externally, this distance must equal the radius of the circle, which we denote as \(r\):
\[ r = |h - 4| \]

5. **Distance from the Center to the Given Circle**:
The distance from the center \((h, k)\) to the center of the given circle \((0, 0)\) is:
\[ \sqrt{h^2 + k^2} \]
Since the circle touches the given circle externally, this distance must equal the sum of the radius of the given circle and the radius of the required circle:
\[ \sqrt{h^2 + k^2} = r + r_1 = r + 2 \]

6. **Substituting for \(r\)**:
From step 4, we have \(r = |h - 4|\). Thus, we can write:
\[ \sqrt{h^2 + k^2} = |h - 4| + 2 \]

7. **Squaring Both Sides**:
To eliminate the square root, we square both sides:
\[ h^2 + k^2 = (|h - 4| + 2)^2 \]
Expanding the right side:
\[ h^2 + k^2 = (h - 4)^2 + 4|h - 4| + 4 \]

8. **Considering Cases for \(|h - 4|\)**:
We need to consider two cases for \(|h - 4|\):
- Case 1: \(h - 4 \geq 0\) (i.e., \(h \geq 4\))
- Case 2: \(h - 4 < 0\) (i.e., \(h < 4\))

**Case 1**: \(h \geq 4\)
\[ h^2 + k^2 = (h - 4)^2 + 4(h - 4) + 4 \]
Simplifying gives:
\[ h^2 + k^2 = h^2 - 8h + 16 + 4h - 16 + 4 \]
\[ k^2 = -4h + 4 \]
\[ k^2 = -4(h - 1) \]

**Case 2**: \(h < 4\)
\[ h^2 + k^2 = (4 - h)^2 + 4(4 - h) + 4 \]
Simplifying gives:
\[ h^2 + k^2 = 16 - 8h + h^2 + 16 - 4h + 4 \]
\[ k^2 = -12h + 36 \]

9. **Finding the Locus**:
The locus equations from both cases can be combined. The locus is a parabola:
\[ k^2 = -12(h - 3) \]
This is in the form \(k^2 = -4p(h - h_0)\) where \(h_0 = 3\) and \(p = 3\).

10. **Distance of the Vertex from the Origin**:
The vertex of the parabola is at \((3, 0)\). The distance from the origin \((0, 0)\) to the vertex \((3, 0)\) is:
\[ \text{Distance} = \sqrt{(3 - 0)^2 + (0 - 0)^2} = \sqrt{9} = 3 \]

### Final Answer:
The distance of the vertex of the locus from the origin is **3**.