Two perpendicular tangents to the circle `x^2 + y^2= a^2` meet at P. Then the locus of P has the equation

Options

A. `x^(2)+y^(2)=2a^(2)`
B. `x^(2)+y^(2)=3a^(2)`
C. `x^(2)+y^(2)=4a^(2)`
D. `x^(2)+y^(2)=5a^(2)`

Answer

To find the locus of the point P where two perpendicular tangents to the circle \(x^2 + y^2 = a^2\) meet, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Circle Equation**:
The given circle is defined by the equation:
\[ x^2 + y^2 = a^2 \]

2. **Assume Coordinates of P**:
Let the coordinates of point P be \(P(h, k)\).

3. **Equation of Tangent to the Circle**:
The equation of the tangent to the circle at a point can be expressed as:
\[ y = mx + c \sqrt{1 + m^2} \]
where \(m\) is the slope of the tangent line.

4. **Substituting Point P into the Tangent Equation**:
Since point P lies on the tangent, we can substitute \(x = h\) and \(y = k\):
\[ k = mh + c \sqrt{1 + m^2} \]
Rearranging gives:
\[ k - mh = c \sqrt{1 + m^2} \]

5. **Squaring Both Sides**:
To eliminate the square root, we square both sides:
\[ (k - mh)^2 = c^2(1 + m^2) \]
Expanding both sides results in:
\[ k^2 - 2kmh + m^2h^2 = c^2 + c^2m^2 \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ k^2 + m^2h^2 - c^2m^2 - 2kmh - c^2 = 0 \]

7. **Collecting Terms**:
We can factor out \(m^2\):
\[ (h^2 - c^2)m^2 - 2kmh + (k^2 - c^2) = 0 \]

8. **Condition for Perpendicular Tangents**:
For the tangents to be perpendicular, the product of their slopes must equal -1:
\[ m_1 \cdot m_2 = -1 \]
This implies that the discriminant of the quadratic in \(m\) must be zero:
\[ (-2kh)^2 - 4(h^2 - c^2)(k^2 - c^2) = 0 \]

9. **Solving the Discriminant**:
Simplifying gives:
\[ 4k^2h^2 = 4(h^2 - c^2)(k^2 - c^2) \]
Dividing by 4 and rearranging leads to:
\[ k^2(h^2 - c^2) = h^2c^2 \]

10. **Substituting \(c^2 = a^2\)**:
Since \(c\) is the distance from the center to the tangent point, we substitute \(c^2 = a^2\):
\[ k^2(h^2 - a^2) = h^2a^2 \]

11. **Final Locus Equation**:
Replacing \(h\) with \(x\) and \(k\) with \(y\) gives:
\[ y^2 = 2a^2 - x^2 \]
Rearranging leads to the final locus equation:
\[ x^2 + y^2 = 2a^2 \]

### Final Answer:
The locus of point P is given by the equation:
\[ x^2 + y^2 = 2a^2 \]