The condition that the circles `x^(2)+y^(2)+2ax+c=0, x^(2)+y^(2)+2by+c=0` may touch each other is

Answer

To find the condition under which the circles given by the equations \(x^2 + y^2 + 2ax + c = 0\) and \(x^2 + y^2 + 2by + c = 0\) touch each other, we will follow these steps:

### Step 1: Identify the centers and radii of the circles

1. **Circle 1**: The equation is \(x^2 + y^2 + 2ax + c = 0\).
- Rearranging gives: \(x^2 + 2ax + y^2 + c = 0\).
- The center \((h_1, k_1)\) is given by \((-a, 0)\).
- The radius \(R_1\) is calculated as:
\[ R_1 = \sqrt{g^2 + f^2 - c} = \sqrt{a^2 - c} \]

2. **Circle 2**: The equation is \(x^2 + y^2 + 2by + c = 0\).
- Rearranging gives: \(x^2 + y^2 + 2by + c = 0\).
- The center \((h_2, k_2)\) is given by \((0, -b)\).
- The radius \(R_2\) is calculated as:
\[ R_2 = \sqrt{g^2 + f^2 - c} = \sqrt{b^2 - c} \]

### Step 2: Calculate the distance between the centers of the circles

The distance \(d\) between the centers \((-a, 0)\) and \((0, -b)\) is given by:
\[ d = \sqrt{(-a - 0)^2 + (0 - (-b))^2} = \sqrt{a^2 + b^2} \]

### Step 3: Set up the condition for the circles to touch each other

For the circles to touch each other, the distance between their centers must equal the sum of their radii:
\[ d = R_1 + R_2 \]
Substituting the expressions we found:
\[ \sqrt{a^2 + b^2} = \sqrt{a^2 - c} + \sqrt{b^2 - c} \]

### Step 4: Square both sides to eliminate the square roots

Squaring both sides gives:
\[ a^2 + b^2 = (R_1 + R_2)^2 = (R_1^2 + R_2^2 + 2R_1R_2) \]
Substituting \(R_1\) and \(R_2\):
\[ a^2 + b^2 = (a^2 - c) + (b^2 - c) + 2\sqrt{(a^2 - c)(b^2 - c)} \]
This simplifies to:
\[ a^2 + b^2 = a^2 + b^2 - 2c + 2\sqrt{(a^2 - c)(b^2 - c)} \]

### Step 5: Rearranging the equation

Rearranging gives:
\[ 2c = 2\sqrt{(a^2 - c)(b^2 - c)} \]
Dividing by 2:
\[ c = \sqrt{(a^2 - c)(b^2 - c)} \]

### Step 6: Square both sides again

Squaring both sides again gives:
\[ c^2 = (a^2 - c)(b^2 - c) \]
Expanding this:
\[ c^2 = a^2b^2 - a^2c - b^2c + c^2 \]
Cancelling \(c^2\) from both sides:
\[ 0 = a^2b^2 - a^2c - b^2c \]
Rearranging gives:
\[ a^2c + b^2c = a^2b^2 \]
Factoring out \(c\):
\[ c(a^2 + b^2) = a^2b^2 \]

### Step 7: Final condition

Thus, the condition for the circles to touch each other is:
\[ \frac{1}{c} = \frac{1}{a^2} + \frac{1}{b^2} \]