A chord through P cut the circle `x^(2)+y^(2)+2gx+2fy+c=0` in A and B another chord through P in c and D, then

Options

A. `PA.PBltPC.PD`
B. `PA.PB=PC.PD`
C. `PA.PC=PB.PD`
D. `PA.PBgtPC.PD`

Answer

To solve the problem, we need to analyze the situation where we have a circle defined by the equation \(x^2 + y^2 + 2gx + 2fy + c = 0\) and two chords through a point \(P\) that intersect the circle at points \(A\), \(B\) and \(C\), \(D\) respectively.

### Step-by-Step Solution:

1. **Identify the Circle and Point P:**
The given circle can be represented as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
Let the coordinates of point \(P\) be \((x_1, y_1)\).

2. **Equation of the Chord through P:**
A chord through point \(P\) can be represented by a line making an angle \(\theta\) with the positive x-axis. The equation of the line through point \(P\) can be expressed as:
\[ y - y_1 = \tan(\theta)(x - x_1) \]
This can be rearranged to:
\[ y = y_1 + \tan(\theta)(x - x_1) \]

3. **Parametric Representation of Points A and B:**
We can express the coordinates of points \(A\) and \(B\) on the chord in terms of a parameter \(L\):
\[ x = x_1 + L \cos(\theta) \]
\[ y = y_1 + L \sin(\theta) \]

4. **Substituting into the Circle Equation:**
Substitute the parametric equations of \(x\) and \(y\) into the circle's equation:
\[ (x_1 + L \cos(\theta))^2 + (y_1 + L \sin(\theta))^2 + 2g(x_1 + L \cos(\theta)) + 2f(y_1 + L \sin(\theta)) + c = 0 \]

5. **Expanding the Equation:**
Expanding the above equation gives:
\[ (x_1^2 + 2x_1L \cos(\theta) + L^2 \cos^2(\theta)) + (y_1^2 + 2y_1L \sin(\theta) + L^2 \sin^2(\theta)) + 2g(x_1 + L \cos(\theta)) + 2f(y_1 + L \sin(\theta)) + c = 0 \]
Collecting like terms leads to a quadratic equation in \(L\).

6. **Identifying the Quadratic Form:**
The quadratic equation in \(L\) can be expressed as:
\[ AL^2 + BL + C = 0 \]
where \(A\), \(B\), and \(C\) are coefficients derived from the expanded equation.

7. **Roots of the Quadratic Equation:**
Let the roots of this quadratic equation be \(R_1\) and \(R_2\) corresponding to points \(A\) and \(B\). The product of the roots is given by:
\[ R_1 \cdot R_2 = \frac{C}{A} \]

8. **Applying the Same for Chord CD:**
Similarly, for the chord through \(P\) intersecting the circle at points \(C\) and \(D\), we can derive another quadratic equation in \(L\) and find the product of its roots \(S_1\) and \(S_2\).

9. **Conclusion:**
Since both products of the roots from the two chords are equal, we conclude:
\[ PA \cdot PB = PC \cdot PD \]

### Final Answer:
Thus, the required relationship is:
\[ PA \cdot PB = PC \cdot PD \]