If the tangent from P to the circle `x^(2)+y^(2)=1` is perpendicular to the tangent from P to the circle `x^(2)+y^(2)=3`, then P lies on a circle of radius 2 concentric with the given circles.

Answer

To solve the problem step by step, we need to analyze the conditions given and derive the required equation for the locus of point P.

### Step-by-Step Solution:

1. **Identify the Circles**:
- The first circle is given by the equation \( x^2 + y^2 = 1 \). This circle has a center at \( C_1(0, 0) \) and a radius \( r_1 = 1 \).
- The second circle is given by the equation \( x^2 + y^2 = 3 \). This circle also has a center at \( C_2(0, 0) \) and a radius \( r_2 = \sqrt{3} \).

2. **Understand the Condition**:
- The problem states that the tangents from point P to both circles are perpendicular. This means that if we denote the angles the tangents make with a reference line (e.g., the x-axis), the sum of these angles should be \( 90^\circ \).

3. **Use the Tangent Length Formula**:
- The length of the tangent from a point \( P(x_1, y_1) \) to a circle \( x^2 + y^2 = r^2 \) is given by \( \sqrt{x_1^2 + y_1^2 - r^2} \).
- Therefore, the lengths of the tangents from P to the two circles are:
- For the first circle: \( L_1 = \sqrt{x_1^2 + y_1^2 - 1} \)
- For the second circle: \( L_2 = \sqrt{x_1^2 + y_1^2 - 3} \)

4. **Condition for Perpendicular Tangents**:
- Since the tangents are perpendicular, we have:
\[ L_1^2 + L_2^2 = (L_1 + L_2)^2 \]
- Expanding this gives:
\[ L_1^2 + L_2^2 = L_1^2 + 2L_1L_2 + L_2^2 \]
- This simplifies to:
\[ 0 = 2L_1L_2 \]
- Since \( L_1 \) and \( L_2 \) cannot be zero (as P is outside both circles), we conclude that \( L_1 \) and \( L_2 \) must satisfy a specific relationship.

5. **Set Up the Relationship**:
- From the lengths we derived, we have:
\[ \sqrt{x_1^2 + y_1^2 - 1} \cdot \sqrt{x_1^2 + y_1^2 - 3} = 0 \]
- This implies that:
\[ x_1^2 + y_1^2 - 1 = 0 \quad \text{or} \quad x_1^2 + y_1^2 - 3 = 0 \]

6. **Find the Locus of Point P**:
- Let \( r^2 = x_1^2 + y_1^2 \). From the above tangents, we can set:
\[ r^2 - 1 = 0 \quad \text{and} \quad r^2 - 3 = 0 \]
- This leads us to find that \( r^2 = 2 \), which means that the locus of point P lies on a circle of radius \( \sqrt{2} \).

7. **Final Equation of the Circle**:
- The equation of the circle with radius 2 and center at the origin is:
\[ x^2 + y^2 = 2^2 \Rightarrow x^2 + y^2 = 4 \]

### Conclusion:
Thus, we have shown that point P lies on a circle of radius 2 concentric with the given circles.