Prove that the radii of the circles `x^(2)+y^(2)=1,x^(2)+y^(2)-2x-6y=6andx^(2)+y^(2)-4x-12y=9` are in AP.

Answer

To prove that the radii of the circles given by the equations \( x^2 + y^2 = 1 \), \( x^2 + y^2 - 2x - 6y = 6 \), and \( x^2 + y^2 - 4x - 12y = 9 \) are in arithmetic progression (AP), we will follow these steps:

### Step 1: Identify the equations of the circles
The equations of the circles are:
1. \( C_1: x^2 + y^2 = 1 \)
2. \( C_2: x^2 + y^2 - 2x - 6y = 6 \)
3. \( C_3: x^2 + y^2 - 4x - 12y = 9 \)

### Step 2: Rewrite the equations in standard form
For a circle in the form \( x^2 + y^2 + 2gx + 2fy + c = 0 \), the radius \( r \) can be calculated using the formula:
\[ r = \sqrt{g^2 + f^2 - c} \]

### Step 3: Find the radius of the first circle \( C_1 \)
For \( C_1: x^2 + y^2 - 1 = 0 \):
- Here, \( g = 0 \), \( f = 0 \), and \( c = -1 \).
- The radius \( r_1 \) is:
\[ r_1 = \sqrt{0^2 + 0^2 - (-1)} = \sqrt{1} = 1 \]

### Step 4: Find the radius of the second circle \( C_2 \)
For \( C_2: x^2 + y^2 - 2x - 6y - 6 = 0 \):
- Here, \( g = -1 \), \( f = -3 \), and \( c = -6 \).
- The radius \( r_2 \) is:
\[ r_2 = \sqrt{(-1)^2 + (-3)^2 - (-6)} = \sqrt{1 + 9 + 6} = \sqrt{16} = 4 \]

### Step 5: Find the radius of the third circle \( C_3 \)
For \( C_3: x^2 + y^2 - 4x - 12y - 9 = 0 \):
- Here, \( g = -2 \), \( f = -6 \), and \( c = -9 \).
- The radius \( r_3 \) is:
\[ r_3 = \sqrt{(-2)^2 + (-6)^2 - (-9)} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7 \]

### Step 6: Check if the radii are in arithmetic progression
The radii we found are:
- \( r_1 = 1 \)
- \( r_2 = 4 \)
- \( r_3 = 7 \)

To check if they are in AP, we need to verify if:
\[ 2r_2 = r_1 + r_3 \]
Calculating:
\[ 2 \cdot 4 = 8 \quad \text{and} \quad 1 + 7 = 8 \]
Since both sides are equal, the radii are in arithmetic progression.

### Conclusion
The radii of the circles \( r_1 = 1 \), \( r_2 = 4 \), and \( r_3 = 7 \) are in arithmetic progression.

---