If the tangent from a point P to the circle `x^(2)+y^(2) = 1` is perpendicular to the tangent from P to `x^(2)+y^(2)=3` then the locus of P is a circle of radius

Options

A. 4
B. 3
C. 2
D. none

Answer

To solve the problem, we need to find the locus of the point \( P \) from which tangents are drawn to two circles such that the tangents are perpendicular to each other. The circles given are:

1. Circle 1: \( x^2 + y^2 = 1 \) (radius \( r_1 = 1 \))
2. Circle 2: \( x^2 + y^2 = 3 \) (radius \( r_2 = \sqrt{3} \))

### Step-by-Step Solution:

1. **Understanding the Tangents**:
The tangents from point \( P(x, y) \) to the circles can be represented using the formula for the length of the tangent from a point to a circle. For a circle \( x^2 + y^2 = r^2 \), the length of the tangent from point \( P(x, y) \) is given by:
\[ L = \sqrt{x^2 + y^2 - r^2} \]

2. **Tangent Lengths**:
- For Circle 1 (\( r_1 = 1 \)):
\[ L_1 = \sqrt{x^2 + y^2 - 1} \]
- For Circle 2 (\( r_2 = \sqrt{3} \)):
\[ L_2 = \sqrt{x^2 + y^2 - 3} \]

3. **Condition for Perpendicular Tangents**:
The tangents from point \( P \) to the circles are perpendicular if the product of their lengths is equal to the square of the radius of the circles:
\[ L_1^2 + L_2^2 = (r_1 + r_2)^2 \]
In our case:
\[ L_1^2 + L_2^2 = 1 + 3 = 4 \]

4. **Setting Up the Equation**:
We can now set up the equation using the lengths of the tangents:
\[ (x^2 + y^2 - 1) + (x^2 + y^2 - 3) = 4 \]
Simplifying this gives:
\[ 2(x^2 + y^2) - 4 = 4 \]
\[ 2(x^2 + y^2) = 8 \]
\[ x^2 + y^2 = 4 \]

5. **Conclusion**:
The locus of point \( P \) is a circle centered at the origin with a radius of \( 2 \).

### Final Answer:
The locus of \( P \) is a circle of radius \( 2 \).