The angle between the pair of tangents from the point `(1, 1/2)` to the circle `x^2 + y^2 + 4x + 2y -4 = 0` is

Options

A. `cos^(-1).(4)/(5)`
B. `sin^(-1).(4)/(5)`
C. `sin^(-1).(3)/(5)`
D. none of these

Answer

To find the angle between the pair of tangents from the point \( (1, \frac{1}{2}) \) to the circle given by the equation \( x^2 + y^2 + 4x + 2y - 4 = 0 \), we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 + 4x + 2y - 4 = 0 \]
We can complete the square for both \( x \) and \( y \).

For \( x \):
\[ x^2 + 4x = (x + 2)^2 - 4 \]

For \( y \):
\[ y^2 + 2y = (y + 1)^2 - 1 \]

Substituting these into the circle equation:
\[ (x + 2)^2 - 4 + (y + 1)^2 - 1 - 4 = 0 \]
\[ (x + 2)^2 + (y + 1)^2 - 9 = 0 \]
\[ (x + 2)^2 + (y + 1)^2 = 9 \]

This shows that the center of the circle is \( (-2, -1) \) and the radius is \( 3 \).

### Step 2: Calculate the Length of the Tangent
The length of the tangent from a point \( (x_1, y_1) \) to a circle with center \( (h, k) \) and radius \( r \) can be calculated using the formula:
\[ L = \sqrt{(x_1 - h)^2 + (y_1 - k)^2 - r^2} \]

Here, \( (x_1, y_1) = (1, \frac{1}{2}) \), \( (h, k) = (-2, -1) \), and \( r = 3 \).

Calculating \( L \):
\[ L = \sqrt{(1 - (-2))^2 + \left(\frac{1}{2} - (-1)\right)^2 - 3^2} \]
\[ = \sqrt{(1 + 2)^2 + \left(\frac{1}{2} + 1\right)^2 - 9} \]
\[ = \sqrt{3^2 + \left(\frac{3}{2}\right)^2 - 9} \]
\[ = \sqrt{9 + \frac{9}{4} - 9} \]
\[ = \sqrt{\frac{9}{4}} = \frac{3}{2} \]

### Step 3: Use the Tangent Length to Find the Angle
The angle \( \theta \) between the tangents can be found using the relationship:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{L}{r} \]
Substituting \( L = \frac{3}{2} \) and \( r = 3 \):
\[ \tan\left(\frac{\theta}{2}\right) = \frac{\frac{3}{2}}{3} = \frac{1}{2} \]

### Step 4: Calculate \( \theta \)
Using the double angle formula for tangent:
\[ \tan \theta = \frac{2 \tan\left(\frac{\theta}{2}\right)}{1 - \tan^2\left(\frac{\theta}{2}\right)} \]
Substituting \( \tan\left(\frac{\theta}{2}\right) = \frac{1}{2} \):
\[ \tan \theta = \frac{2 \cdot \frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]

### Step 5: Find the Angle \( \theta \)
To find \( \theta \), we can use the inverse tangent:
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]

### Conclusion
Thus, the angle between the pair of tangents from the point \( (1, \frac{1}{2}) \) to the circle is \( \tan^{-1}\left(\frac{4}{3}\right) \).