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

Options

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

Answer

To determine 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 can follow these steps:

### Step 1: Rewrite the Circle Equations
The given equations can be rewritten in standard form.

1. For the first circle:
\[ x^2 + y^2 + 2ax + c = 0 \implies (x + a)^2 + y^2 = a^2 - c \]
Here, the center is \((-a, 0)\) and the radius \(r_1 = \sqrt{a^2 - c}\).

2. For the second circle:
\[ x^2 + y^2 + 2by + c = 0 \implies x^2 + (y + b)^2 = b^2 - c \]
Here, the center is \((0, -b)\) and the radius \(r_2 = \sqrt{b^2 - c}\).

### Step 2: Distance Between the Centers
Next, we find the distance \(d\) between the centers of the two circles:
\[ d = \sqrt{((-a) - 0)^2 + (0 - (-b))^2} = \sqrt{a^2 + b^2} \]

### Step 3: Condition for Circles to Touch
For the circles to touch each other externally, 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, we square both sides:
\[ a^2 + b^2 = ( \sqrt{a^2 - c} + \sqrt{b^2 - c} )^2 \]
Expanding the right-hand 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:
\[ 2c = 2\sqrt{(a^2 - c)(b^2 - c)} \]
Dividing both sides by 2:
\[ c = \sqrt{(a^2 - c)(b^2 - c)} \]

### Step 6: Square Again
Squaring both sides again:
\[ c^2 = (a^2 - c)(b^2 - c) \]
Expanding the right-hand 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 the Final Condition
Rearranging gives:
\[ a^2b^2 = a^2c + b^2c \]
Dividing through by \(c^2\):
\[ \frac{1}{c^2} = \frac{1}{a^2} + \frac{1}{b^2} \]

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