A point A(2, 1) is outside the circle `x^2+y^2+2gx+2fy+c=0` & AP, AQ are tangents to the circle.The equation of the circle circumscribing the triangle APQ is:

Options

A. `(x+g)(x-2)+(y+f)(y-1)=0`
B. `(x+g)(x-2)-(y+f)(y-1)=0`
C. `(x-g)(x+2)+(y-f)(y+1)=0`
D. `(x+g)(x+2)+(y+f)(y+1)=0`

Explanation

To find the equation of the circle that circumscribes triangle APQ, we will follow these steps:

### Step 1: Understand the Given Information
We have a point \( A(2, 1) \) outside the circle given by the equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \). The points \( P \) and \( Q \) are the points of tangency from point \( A \) to the circle.

### Step 2: Identify the Center of the Circle
The center \( O \) of the circle can be found from the standard form of the circle equation. The center is given by the coordinates \( (-g, -f) \).

### Step 3: Determine the Slopes of Tangents
Let the coordinates of points \( P \) and \( Q \) be \( (h, k) \). The slope of line segment \( AP \) is given by:
\[ m_1 = \frac{k - 1}{h - 2} \]
The slope of line segment \( OP \) is:
\[ m_2 = \frac{k + f}{h + g} \]

### Step 4: Use the Perpendicularity Condition
Since \( AP \) is tangent to the circle at point \( P \), the slopes \( m_1 \) and \( m_2 \) are perpendicular. Therefore, we have:
\[ m_1 \cdot m_2 = -1 \]
Substituting the values of \( m_1 \) and \( m_2 \):
\[ \left(\frac{k - 1}{h - 2}\right) \cdot \left(\frac{k + f}{h + g}\right) = -1 \]

### Step 5: Simplify the Equation
Cross-multiplying gives:
\[ (k - 1)(k + f) = -(h - 2)(h + g) \]
Expanding both sides:
\[ k^2 + fk - k - f = -h^2 - gh + 2h + 2g \]

### Step 6: Rearranging the Equation
Rearranging the equation leads to:
\[ k^2 + (f + 1)k + (h^2 + gh - 2h - 2g - f) = 0 \]

### Step 7: Locus of Point A
To find the locus of point \( A \) as it varies, we replace \( h \) with \( x \) and \( k \) with \( y \):
\[ y^2 + (f + 1)y + (x^2 + gx - 2x - 2g - f) = 0 \]

### Step 8: Final Equation
This represents the equation of the circle that circumscribes triangle \( APQ \). The final equation can be simplified and rearranged as needed.

### Conclusion
The equation of the circle that circumscribes triangle \( APQ \) is:
\[ x^2 + y^2 + (g + 2)x + (f + 1)y + (c + 2g) = 0 \]