If two circles `x^2+y^2+2gx+2fy=0` and `x^2+y^2+2g'x+2f'y=0` touch each other , then `((f')/(f))(g/(g'))`=___

Answer

To solve the problem, we need to find the value of \(\frac{f'}{f} \cdot \frac{g}{g'}\) given that two circles touch each other.

### Step-by-Step Solution:

1. **Identify the equations of the circles**:
The equations of the circles are given as:
\[ x^2 + y^2 + 2gx + 2fy = 0 \quad \text{(Circle 1)} \]
\[ x^2 + y^2 + 2g'x + 2f'y = 0 \quad \text{(Circle 2)} \]

2. **Find the centers and radii of the circles**:
For Circle 1, the center is at \((-g, -f)\) and the radius \(R_1\) is given by:
\[ R_1 = \sqrt{g^2 + f^2} \]
For Circle 2, the center is at \((-g', -f')\) and the radius \(R_2\) is:
\[ R_2 = \sqrt{g'^2 + f'^2} \]

3. **Condition for the circles to touch**:
The circles can touch each other either externally or internally. The conditions are:
- Externally: \(c_1c_2 = R_1 + R_2\)
- Internally: \(c_1c_2 = |R_1 - R_2|\)

Here, \(c_1c_2\) is the distance between the centers of the circles:
\[ c_1c_2 = \sqrt{(-g + g')^2 + (-f + f')^2} \]

4. **Set up the equation for external touching**:
For external touching, we have:
\[ \sqrt{(-g + g')^2 + (-f + f')^2} = \sqrt{g^2 + f^2} + \sqrt{g'^2 + f'^2} \]
Squaring both sides gives:
\[ (-g + g')^2 + (-f + f')^2 = (g^2 + f^2) + (g'^2 + f'^2) + 2\sqrt{(g^2 + f^2)(g'^2 + f'^2)} \]

5. **Simplify the equation**:
Expanding both sides leads to:
\[ g^2 - 2gg' + g'^2 + f^2 - 2ff' + f'^2 = g^2 + f^2 + g'^2 + f'^2 + 2\sqrt{(g^2 + f^2)(g'^2 + f'^2)} \]
Canceling \(g^2 + f^2 + g'^2 + f'^2\) from both sides results in:
\[ -2gg' - 2ff' = 2\sqrt{(g^2 + f^2)(g'^2 + f'^2)} \]

6. **Rearranging terms**:
Dividing by 2 gives:
\[ -gg' - ff' = \sqrt{(g^2 + f^2)(g'^2 + f'^2)} \]

7. **Final relationship**:
From the above equation, we can derive:
\[ (gf' - fg')^2 = 0 \]
This implies:
\[ gf' = fg' \]
Therefore, we can express this as:
\[ \frac{f'}{f} = \frac{g}{g'} \]

8. **Conclusion**:
Thus, we find:
\[ \frac{f'}{f} \cdot \frac{g}{g'} = 1 \]

### Final Answer:
\[ \frac{f'}{f} \cdot \frac{g}{g'} = 1 \]