The locus of the point of intersection of tangents to the circle `x^(2)+y^(2)=a^(2)` at the points whose parametric angles differ by `pi//3` is

Options

A. `x^(2)+y^(2)=a^(2)`
B. `x^(2)+y^(2)=4a^(2)`
C. `x^(2)+y^(2)=4a^(2)//3`
D. `x^(2)+y^(2)=9`

Answer

To find the locus of the point of intersection of tangents to the circle \(x^2 + y^2 = a^2\) at points whose parametric angles differ by \(\frac{\pi}{3}\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Circle and Points**:
The given circle is \(x^2 + y^2 = a^2\). The center of the circle is at the origin (0, 0) and the radius is \(a\). Let \(P\) be a point on the circle corresponding to the angle \(\theta\):
\[ P(a \cos \theta, a \sin \theta) \]

2. **Find the Second Point**:
The second point \(Q\) is at an angle \(\theta + \frac{\pi}{3}\):
\[ Q\left(a \cos\left(\theta + \frac{\pi}{3}\right), a \sin\left(\theta + \frac{\pi}{3}\right)\right) \]

3. **Parametric Coordinates**:
Using the angle addition formulas:
\[ Q\left(a \left(\cos \theta \cdot \frac{1}{2} - \sin \theta \cdot \frac{\sqrt{3}}{2}\right), a \left(\sin \theta \cdot \frac{1}{2} + \cos \theta \cdot \frac{\sqrt{3}}{2}\right)\right) \]

4. **Equation of Tangents**:
The equation of the tangent to the circle at point \(P\) is given by:
\[ x \cdot (a \cos \theta) + y \cdot (a \sin \theta) = a^2 \]
Simplifying gives:
\[ x \cos \theta + y \sin \theta = a \]

For point \(Q\):
\[ x \cdot \left(a \left(\frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta\right)\right) + y \cdot \left(a \left(\frac{1}{2} \sin \theta + \frac{\sqrt{3}}{2} \cos \theta\right)\right) = a^2 \]

5. **Simplifying the Tangent Equations**:
The tangent equations can be simplified and combined. After some algebra, we can express the equations in a more manageable form.

6. **Finding the Locus**:
By manipulating the equations, we can derive the relationship:
\[ x^2 + y^2 = \frac{4a^2}{3} \]

### Final Result:
Thus, the locus of the point of intersection of the tangents is:
\[ x^2 + y^2 = \frac{4a^2}{3} \]