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

Answer

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

### Step 1: Rewrite the Circle Equations
We start with the given equations of the circles:
1. \(x^2 + y^2 + 2ax + c^2 = 0\)
2. \(x^2 + y^2 + 2by + c^2 = 0\)

To rewrite these in standard form, we complete the square for both equations.

### Step 2: Completing the Square for the First Circle
For the first circle:
\[ x^2 + 2ax + y^2 + c^2 = 0 \]
Completing the square for \(x\):
\[ (x + a)^2 - a^2 + y^2 + c^2 = 0 \]
Rearranging gives:
\[ (x + a)^2 + y^2 = a^2 - c^2 \]
This represents a circle with center \((-a, 0)\) and radius \(R_1 = \sqrt{a^2 - c^2}\).

### Step 3: Completing the Square for the Second Circle
For the second circle:
\[ x^2 + y^2 + 2by + c^2 = 0 \]
Completing the square for \(y\):
\[ x^2 + (y + b)^2 - b^2 + c^2 = 0 \]
Rearranging gives:
\[ x^2 + (y + b)^2 = b^2 - c^2 \]
This represents a circle with center \((0, -b)\) and radius \(R_2 = \sqrt{b^2 - c^2}\).

### Step 4: Finding the Distance Between the Centers
The centers of the circles are:
- Center of Circle 1: \(C_1(-a, 0)\)
- Center of Circle 2: \(C_2(0, -b)\)

The distance \(d\) between the centers \(C_1\) and \(C_2\) is given by:
\[ d = \sqrt{(-a - 0)^2 + (0 - (-b))^2} = \sqrt{a^2 + b^2} \]

### Step 5: Condition for the Circles to Touch
For the circles to touch each other, the distance between the centers must equal the sum of the radii:
\[ \sqrt{a^2 + b^2} = R_1 + R_2 \]
Substituting the radii:
\[ \sqrt{a^2 + b^2} = \sqrt{a^2 - c^2} + \sqrt{b^2 - c^2} \]

### Step 6: Squaring Both Sides
Squaring both sides gives:
\[ a^2 + b^2 = (R_1 + R_2)^2 \]
Expanding the right side:
\[ a^2 + b^2 = ( \sqrt{a^2 - c^2} + \sqrt{b^2 - c^2} )^2 \]
\[ = (a^2 - c^2) + (b^2 - c^2) + 2\sqrt{(a^2 - c^2)(b^2 - c^2)} \]
This simplifies to:
\[ a^2 + b^2 = a^2 + b^2 - 2c^2 + 2\sqrt{(a^2 - c^2)(b^2 - c^2)} \]
Cancelling \(a^2 + b^2\) from both sides leads to:
\[ 0 = -2c^2 + 2\sqrt{(a^2 - c^2)(b^2 - c^2)} \]
Thus,
\[ 2c^2 = 2\sqrt{(a^2 - c^2)(b^2 - c^2)} \]
Dividing by 2:
\[ c^2 = \sqrt{(a^2 - c^2)(b^2 - c^2)} \]
Squaring both sides again gives:
\[ c^4 = (a^2 - c^2)(b^2 - c^2) \]
Expanding the right side:
\[ c^4 = a^2b^2 - a^2c^2 - b^2c^2 + c^4 \]
Cancelling \(c^4\) from both sides results in:
\[ 0 = a^2b^2 - a^2c^2 - b^2c^2 \]
Rearranging gives:
\[ a^2c^2 + b^2c^2 = a^2b^2 \]
Dividing by \(a^2b^2c^2\) leads to:
\[ \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2} \]
Thus, we have proved that the circles touch each other under the given condition.