If `alpha` is the angle subtended at `P(x_(1),y_(1))` by the circle `S-=x^(2)+y^(2)+2gx+2fy+c=0` then

Options

A. `cotalpha=(sqrtS_(1))/(sqrt((g^(2)+f^(2)-c)))`
B. `cot""(alpha)/(2)=(sqrtS_(1))/(sqrt((g^(2)+f^(2)-c)))`
C. `tanalpha=(2sqrt((g^(2)+f^(2)-c)))/(sqrtS_(1))`
D. `alpha=2tan^(-1)((sqrt((g^(2)+f^(2)-c)))/(sqrtS_(1)))`

Answer

To solve the problem, we need to find the angle \( \alpha \) subtended at the point \( P(x_1, y_1) \) by the circle given by the equation \( S = x^2 + y^2 + 2gx + 2fy + c = 0 \).

### Step-by-Step Solution:

1. **Identify the Circle and Point**:
The equation of the circle is given as:
\[ S = x^2 + y^2 + 2gx + 2fy + c = 0 \]
The center of the circle \( C \) is at \( (-g, -f) \) and the radius \( R \) is given by:
\[ R = \sqrt{g^2 + f^2 - c} \]

2. **Calculate \( S_1 \)**:
We need to evaluate \( S_1 \) at the point \( P(x_1, y_1) \):
\[ S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c \]

3. **Tangent Length from Point to Circle**:
The length of the tangent \( PA \) from point \( P \) to the circle can be calculated using the formula:
\[ PA = \sqrt{S_1} \]

4. **Distance from Point to Center**:
The distance \( AC \) from point \( P \) to the center \( C(-g, -f) \) is given by:
\[ AC = \sqrt{(x_1 + g)^2 + (y_1 + f)^2} \]

5. **Using the Tangent-Secant Theorem**:
The angle \( \alpha \) subtended at point \( P \) by the tangents \( PA \) and \( PB \) can be related to the lengths \( PA \) and \( AC \). We can use the cotangent of half the angle:
\[ \cot\left(\frac{\alpha}{2}\right) = \frac{PA}{AC} \]

6. **Substituting Values**:
Substituting the values we found:
\[ \cot\left(\frac{\alpha}{2}\right) = \frac{\sqrt{S_1}}{\sqrt{g^2 + f^2 - c}} \]

7. **Finding \( \tan\left(\frac{\alpha}{2}\right) \)**:
From the cotangent relation, we can find \( \tan\left(\frac{\alpha}{2}\right) \):
\[ \tan\left(\frac{\alpha}{2}\right) = \frac{\sqrt{g^2 + f^2 - c}}{\sqrt{S_1}} \]

8. **Finding \( \alpha \)**:
Finally, we can express \( \alpha \) in terms of the tangent:
\[ \alpha = 2 \tan^{-1}\left(\frac{\sqrt{g^2 + f^2 - c}}{\sqrt{S_1}}\right) \]

### Conclusion:
The angle \( \alpha \) subtended at point \( P(x_1, y_1) \) by the circle is given by:
\[ \alpha = 2 \tan^{-1}\left(\frac{\sqrt{g^2 + f^2 - c}}{\sqrt{S_1}}\right) \]