Consider the circles `x^(2) + y^(2) + 2ax + c = 0` and `x^(2) + y^(2) + 2by + c = 0`
The two circles touch each other . If :

Options

A. `c = sqrt(a^(2) + b^(2))`
B. `(1)/ (c )= (1)/(a^(2)) + (1)/(b^(2))`
C. `c = (1)/(a^(2)) + (1)/(b^(2))`
D. `c = (1)/(a^(2)) + (1)/(b^(2))`

Answer

To solve the problem regarding the two circles given by the equations \(x^2 + y^2 + 2ax + c = 0\) and \(x^2 + y^2 + 2by + c = 0\) that touch each other, we can follow these steps:

### Step 1: Identify the centers and radii of the circles
The general equation of a circle is given by:
\[ x^2 + y^2 + 2Ax + 2By + C = 0 \]
From the first circle equation \(x^2 + y^2 + 2ax + c = 0\), we can identify:
- Center: \((-a, 0)\)
- Radius: \(R_1 = \sqrt{a^2 - c}\)

From the second circle equation \(x^2 + y^2 + 2by + c = 0\), we can identify:
- Center: \((0, -b)\)
- Radius: \(R_2 = \sqrt{b^2 - c}\)

### Step 2: Calculate the distance between the centers
The distance \(d\) between the centers \((-a, 0)\) and \((0, -b)\) can be calculated using the distance formula:
\[ d = \sqrt{((-a) - 0)^2 + (0 - (-b))^2} = \sqrt{a^2 + b^2} \]

### Step 3: Set up the condition for the circles to touch each other
For the two circles to touch each other, the distance between their centers must equal the sum of their radii:
\[ d = R_1 + R_2 \]
Substituting the values 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:
\[ a^2 + b^2 = (R_1 + R_2)^2 = ( \sqrt{a^2 - c} + \sqrt{b^2 - c} )^2 \]
Expanding the right side:
\[ a^2 + b^2 = (a^2 - c) + (b^2 - c) + 2\sqrt{(a^2 - c)(b^2 - c)} \]
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:
\[ 0 = -2c + 2\sqrt{(a^2 - c)(b^2 - c)} \]
Dividing through by 2:
\[ c = \sqrt{(a^2 - c)(b^2 - c)} \]

### Step 6: Square both sides again
Squaring both sides again:
\[ c^2 = (a^2 - c)(b^2 - c) \]
Expanding the right side:
\[ c^2 = a^2b^2 - a^2c - b^2c + c^2 \]
Cancelling \(c^2\) from both sides:
\[ 0 = a^2b^2 - a^2c - b^2c \]

### Step 7: Rearranging to find the condition
Rearranging gives:
\[ a^2c + b^2c = a^2b^2 \]
Factoring out \(c\):
\[ c(a^2 + b^2) = a^2b^2 \]
Thus, we find:
\[ c = \frac{a^2b^2}{a^2 + b^2} \]

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