The angle between the tangents drawn from the origin to the circle `x^(2) + y^(2) + 4x - 6y + 4 = 0` is

Options

A. `tan^(-1)(5/13)`
B. `tan^(-1)(5/12)`
C. `tan^(-1)(12/5)`
D. `tan^(-1)(13/5)`

Answer

To find the angle between the tangents drawn from the origin to the circle given by the equation \( x^2 + y^2 + 4x - 6y + 4 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation of the circle in standard form
The general form of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From the given equation \( x^2 + y^2 + 4x - 6y + 4 = 0 \), we can identify:
- \( 2g = 4 \) which gives \( g = 2 \)
- \( 2f = -6 \) which gives \( f = -3 \)
- \( c = 4 \)

### Step 2: Find the center and radius of the circle
The center of the circle is given by the coordinates \( (-g, -f) \):
\[ \text{Center} = (-2, 3) \]
The radius \( r \) can be calculated using the formula:
\[ r = \sqrt{g^2 + f^2 - c} \]
Substituting the values:
\[ r = \sqrt{2^2 + (-3)^2 - 4} = \sqrt{4 + 9 - 4} = \sqrt{9} = 3 \]

### Step 3: Find the distance from the origin to the center of the circle
The distance \( d \) from the origin \( (0, 0) \) to the center \( (-2, 3) \) is calculated using the distance formula:
\[ d = \sqrt{(-2 - 0)^2 + (3 - 0)^2} = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \]

### Step 4: Use the tangent formula to find the angle
The angle \( \theta \) between the tangents from a point outside the circle can be found using the formula:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{r}{d} \]
Substituting the values of \( r \) and \( d \):
\[ \tan\left(\frac{\theta}{2}\right) = \frac{3}{\sqrt{13}} \]

### Step 5: Find \( \theta \) using the double angle formula
To find \( \theta \), we use the identity:
\[ \tan(\theta) = \frac{2 \tan\left(\frac{\theta}{2}\right)}{1 - \tan^2\left(\frac{\theta}{2}\right)} \]
Substituting \( \tan\left(\frac{\theta}{2}\right) = \frac{3}{\sqrt{13}} \):
\[ \tan(\theta) = \frac{2 \cdot \frac{3}{\sqrt{13}}}{1 - \left(\frac{3}{\sqrt{13}}\right)^2} \]
Calculating \( \left(\frac{3}{\sqrt{13}}\right)^2 = \frac{9}{13} \):
\[ \tan(\theta) = \frac{\frac{6}{\sqrt{13}}}{1 - \frac{9}{13}} = \frac{\frac{6}{\sqrt{13}}}{\frac{4}{13}} = \frac{6 \cdot 13}{4 \cdot \sqrt{13}} = \frac{78}{4\sqrt{13}} = \frac{39}{2\sqrt{13}} \]

### Step 6: Find the angle \( \theta \)
Thus, the angle \( \theta \) can be expressed as:
\[ \theta = \tan^{-1}\left(\frac{39}{2\sqrt{13}}\right) \]

### Final Answer
The angle between the tangents drawn from the origin to the circle is:
\[ \theta = \tan^{-1}\left(\frac{39}{2\sqrt{13}}\right) \]