Prove that the circle `x^(2) + y^(2) + 2ax + c^(2) = 0 and x^(2) + y^(2) + 2by + c^(2) = 0` touch each other if
`(1)/(a^(2)) + (1)/(b^(2)) = (1)/(c^(2))` .

Explanation

To prove that the circles given by the equations \(x^2 + y^2 + 2ax + c^2 = 0\) and \(x^2 + y^2 + 2by + c^2 = 0\) touch each other under the condition \(\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}\), we will follow these steps:

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

The first circle can be rewritten in standard form:
\[ x^2 + y^2 + 2ax + c^2 = 0 \implies (x + a)^2 + y^2 = a^2 - c^2 \]
...