Prove that the tangents from any point of a fixed circle of co-axial system to two other fixed circles of the system are in a constant ratio.

Answer

To prove that the tangents from any point of a fixed circle of a coaxial system to two other fixed circles of the system are in a constant ratio, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Coaxial Circles**:
- Let the equations of the three coaxial circles be given by:
\[ x^2 + y^2 + 2g_1x + c_1 = 0 \quad (Circle 1) \]
\[ x^2 + y^2 + 2g_2x + c_2 = 0 \quad (Circle 2) \]
\[ x^2 + y^2 + 2g_3x + c_3 = 0 \quad (Circle 3) \]
- Here, \(g_1, g_2, g_3\) are constants related to the circles, and \(c_1, c_2, c_3\) are the constant terms.

2. **Point from which Tangents are Drawn**:
- Let \(P(h, k)\) be a point from which tangents are drawn to the circles.

3. **Equation of Tangents**:
- The equation of the tangent from point \(P(h, k)\) to Circle 1 is given by:
\[ h^2 + k^2 + 2g_1h + c_1 = 0 \]
- Rearranging gives:
\[ h^2 + k^2 = -2g_1h - c_1 \quad (1) \]

4. **Tangent Lengths to Circle 2 and Circle 3**:
- For Circle 2:
\[ h^2 + k^2 + 2g_2h + c_2 = 0 \quad (2) \]
- For Circle 3:
\[ h^2 + k^2 + 2g_3h + c_3 = 0 \quad (3) \]

5. **Substituting Equation (1) into Equations (2) and (3)**:
- Substitute \(h^2 + k^2\) from Equation (1) into Equations (2) and (3):
- For Circle 2:
\[ -2g_1h - c_1 + 2g_2h + c_2 = 0 \]
Simplifying gives:
\[ 2(g_2 - g_1)h + (c_2 - c_1) = 0 \quad (4) \]
- For Circle 3:
\[ -2g_1h - c_1 + 2g_3h + c_3 = 0 \]
Simplifying gives:
\[ 2(g_3 - g_1)h + (c_3 - c_1) = 0 \quad (5) \]

6. **Finding the Ratio of Tangents**:
- The lengths of the tangents \(PQ\) and \(PR\) from point \(P\) to Circle 2 and Circle 3 can be expressed as:
\[ PQ^2 = h^2 + k^2 + 2g_2h + c_2 \]
\[ PR^2 = h^2 + k^2 + 2g_3h + c_3 \]
- Therefore, the ratio of the squares of the tangent lengths is:
\[ \frac{PQ^2}{PR^2} = \frac{h^2 + k^2 + 2g_2h + c_2}{h^2 + k^2 + 2g_3h + c_3} \]
- Substituting \(h^2 + k^2\) from Equation (1) into this ratio gives:
\[ \frac{-2g_1h - c_1 + 2g_2h + c_2}{-2g_1h - c_1 + 2g_3h + c_3} \]
- This simplifies to:
\[ \frac{2(g_2 - g_1)h + (c_2 - c_1)}{2(g_3 - g_1)h + (c_3 - c_1)} \]

7. **Conclusion**:
- Since \(g_1, g_2, g_3, c_1, c_2, c_3\) are constants, the ratio \(\frac{PQ^2}{PR^2}\) is constant for any point \(P\).
- Thus, we have shown that the tangents from any point of a fixed circle of a coaxial system to the other two fixed circles are in a constant ratio.