The locus of the point of intersection of the two tangents drawn to the circle `x^2 + y^2=a^2` which include are angle ` alpha` is

Options

A. 2
B. 4
C. 8
D. 16

Answer

To find the locus of the point of intersection of two tangents drawn to the circle \(x^2 + y^2 = a^2\) that include an angle \(\alpha\), we can follow these steps:

### Step 1: Understand the Circle and Tangents
The given circle is centered at the origin (0, 0) with a radius \(a\). We denote a point \(P(h, k)\) from which two tangents are drawn to the circle, forming an angle \(\alpha\) between them.

**Hint:** Visualize the circle and the point from which the tangents are drawn.

### Step 2: Relationship Between Tangents and Circle
The tangents drawn from point \(P\) to the circle will meet at point \(P\) and form an angle \(\alpha\). The distance from the center of the circle \(C(0, 0)\) to the point \(P(h, k)\) is given by \(PC = \sqrt{h^2 + k^2}\).

**Hint:** Remember that the distance from the center to the point of tangency is crucial for finding the locus.

### Step 3: Use the Angle Bisector Theorem
The angle between the two tangents can be bisected. The angle between the radius to the point of tangency and the tangent line is \(90^\circ\). Therefore, we can use the sine function to relate the angle \(\alpha\) and the distances involved.

Using the sine of half the angle:
\[ \sin\left(\frac{\alpha}{2}\right) = \frac{a}{PC} \]
From this, we can express \(PC\):
\[ PC = \frac{a}{\sin\left(\frac{\alpha}{2}\right)} \]

**Hint:** The sine function relates the angle and the lengths in the right triangle formed by the radius and the tangent.

### Step 4: Find the Radius of the Locus Circle
The distance \(PC\) can also be expressed in terms of the radius \(R\) of the locus circle. Therefore, we have:
\[ R = PC = \frac{a}{\sin\left(\frac{\alpha}{2}\right)} \]

**Hint:** This radius \(R\) will be the radius of the new locus circle.

### Step 5: Write the Equation of the Locus Circle
The locus of point \(P\) will be a circle concentric with the original circle. The equation of the locus circle can be expressed as:
\[ x^2 + y^2 = R^2 \]
Substituting \(R\):
\[ x^2 + y^2 = \left(\frac{a}{\sin\left(\frac{\alpha}{2}\right)}\right)^2 \]
This simplifies to:
\[ x^2 + y^2 = \frac{a^2}{\sin^2\left(\frac{\alpha}{2}\right)} \]

**Hint:** The locus is a circle with a radius that depends on the angle \(\alpha\).

### Final Answer
The locus of the point of intersection of the two tangents drawn to the circle \(x^2 + y^2 = a^2\) that include an angle \(\alpha\) is given by:
\[ x^2 + y^2 = \frac{a^2}{\sin^2\left(\frac{\alpha}{2}\right)} \]