If the tangent from a point 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 the locus of p is

Options

A. 4
B. 3
C. 2
D. None of these

Answer

To solve the problem, we need to find the locus of point P such that the tangent from P to the circle \( x^2 + y^2 = 1 \) is perpendicular to the tangent from P to the circle \( x^2 + y^2 = 3 \).

### Step-by-Step Solution:

1. **Identify the circles and their properties**:
- The first circle is given by the equation \( x^2 + y^2 = 1 \). Its center is at \( (0, 0) \) and its radius \( r_1 = 1 \).
- The second circle is given by the equation \( x^2 + y^2 = 3 \). Its center is also at \( (0, 0) \) and its radius \( r_2 = \sqrt{3} \).

2. **Use the formula for the length of the tangent**:
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:
\[ L = \sqrt{x_1^2 + y_1^2 - r^2} \]
- For the first circle (radius = 1):
\[ L_1 = \sqrt{x^2 + y^2 - 1} \]
- For the second circle (radius = \( \sqrt{3} \)):
\[ L_2 = \sqrt{x^2 + y^2 - 3} \]

3. **Condition for perpendicular tangents**:
The tangents from point P to the circles are perpendicular if the product of their slopes is -1. This can be derived from the geometric properties of tangents. However, in this case, we can also use the property of the lengths of the tangents:
\[ L_1^2 + L_2^2 = d^2 \]
where \( d \) is the distance from point P to the origin (0,0), which is \( \sqrt{x^2 + y^2} \).

4. **Set up the equation**:
From the condition of perpendicularity:
\[ L_1^2 + L_2^2 = L_1^2 + L_2^2 = (x^2 + y^2 - 1) + (x^2 + y^2 - 3) = 2(x^2 + y^2) - 4 \]
Since \( d^2 = x^2 + y^2 \):
\[ 2d^2 - 4 = d^2 \]

5. **Solve for d**:
Rearranging the equation:
\[ 2d^2 - d^2 = 4 \implies d^2 = 4 \]
Thus, we have:
\[ x^2 + y^2 = 4 \]

6. **Conclusion**:
The locus of point P is the equation of a circle centered at the origin with radius 2:
\[ x^2 + y^2 = 4 \]