The length of the tangent drawn from any point on the circle `S=x^(2)+y^(2)+2gx+2fy+c=0` to the circle `S'=x^(2)+y^(2)+2gx+2fy+c'=0` where `c' gt c` is

Options

A. `sqrt(c+c')`
B. `sqrt(c-c')`
C. `sqrt(c'-c)`
D. none

Answer

To find the length of the tangent drawn from any point on the circle \( S: x^2 + y^2 + 2gx + 2fy + c = 0 \) to the circle \( S': x^2 + y^2 + 2gx + 2fy + c' = 0 \) where \( c' > c \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the circles**:
- The first circle \( S \) has the equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \).
- The second circle \( S' \) has the equation \( x^2 + y^2 + 2gx + 2fy + c' = 0 \).

2. **Point on the first circle**:
- Let \( (a, b) \) be a point on the circle \( S \). By substituting \( (a, b) \) into the equation of circle \( S \), we have:
\[ a^2 + b^2 + 2ga + 2fb + c = 0 \]
- Rearranging gives:
\[ a^2 + b^2 + 2ga + 2fb = -c \]

3. **Length of the tangent from point \( (a, b) \) to circle \( S' \)**:
- The formula for the length of the tangent \( l \) from a point \( (x_1, y_1) \) to a circle \( S' \) is given by:
\[ l = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c'} \]
- Substituting \( (a, b) \) into this formula gives:
\[ l = \sqrt{a^2 + b^2 + 2ga + 2fb + c'} \]

4. **Substituting the value from step 2**:
- From step 2, we know that \( a^2 + b^2 + 2ga + 2fb = -c \). Thus, we can substitute this into the tangent length formula:
\[ l = \sqrt{-c + c'} \]
- This simplifies to:
\[ l = \sqrt{c' - c} \]

5. **Final answer**:
- Therefore, the length of the tangent drawn from any point on the circle \( S \) to the circle \( S' \) is:
\[ l = \sqrt{c' - c} \]

### Summary:
The length of the tangent drawn from any point on the circle \( S \) to the circle \( S' \) is \( \sqrt{c' - c} \).