The circle `x^(2)+y^(2)-2ax+c^(2)=0` and `x^(2)+y^(2)-2by+c^(2)=0` will touch each other externally if

Options

A. `(1)/(a^(2)) +(1)/(b^(2)) =(1)/(c^(2))`
B. `(1)/(b^(2)) +(1)/(c^(2))=(1)/(a^(2))`
C. `(1)/(c^(2)) +(1)/(a^(2)) =(1)/(b^(2))`
D. none

Answer

To determine the condition under which 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 externally, we can follow these steps:

### Step 1: Identify the centers and radii of the circles
For the first circle \(S_1: x^2 + y^2 - 2ax + c^2 = 0\):
- The center \(C_1\) is \((a, 0)\).
- The radius \(r_1\) is given by:
\[ r_1 = \sqrt{a^2 - c^2} \]

For the second circle \(S_2: x^2 + y^2 - 2by + c^2 = 0\):
- The center \(C_2\) is \((0, b)\).
- The radius \(r_2\) is given by:
\[ r_2 = \sqrt{b^2 - c^2} \]

### Step 2: Find the distance between the centers
The distance \(d\) between the centers \(C_1\) and \(C_2\) is calculated as:
\[ d = \sqrt{(a - 0)^2 + (0 - b)^2} = \sqrt{a^2 + b^2} \]

### Step 3: Set up the condition for external tangency
For the circles to touch each other externally, the distance between the centers must equal the sum of the radii:
\[ d = r_1 + r_2 \]
Substituting the expressions for \(d\), \(r_1\), and \(r_2\):
\[ \sqrt{a^2 + b^2} = \sqrt{a^2 - c^2} + \sqrt{b^2 - c^2} \]

### Step 4: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ a^2 + b^2 = ( \sqrt{a^2 - c^2} + \sqrt{b^2 - c^2} )^2 \]
Expanding the right-hand side:
\[ a^2 + b^2 = (a^2 - c^2) + (b^2 - c^2) + 2\sqrt{(a^2 - c^2)(b^2 - c^2)} \]
This simplifies to:
\[ a^2 + b^2 = a^2 + b^2 - 2c^2 + 2\sqrt{(a^2 - c^2)(b^2 - c^2)} \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ 0 = -2c^2 + 2\sqrt{(a^2 - c^2)(b^2 - c^2)} \]
Dividing by 2:
\[ c^2 = \sqrt{(a^2 - c^2)(b^2 - c^2)} \]

### Step 6: Square both sides again
Squaring both sides again results in:
\[ c^4 = (a^2 - c^2)(b^2 - c^2) \]
Expanding the right-hand side:
\[ c^4 = a^2b^2 - a^2c^2 - b^2c^2 + c^4 \]
Subtracting \(c^4\) from both sides gives:
\[ 0 = a^2b^2 - a^2c^2 - b^2c^2 \]

### Step 7: Factor the equation
Factoring out \(c^2\):
\[ a^2b^2 = c^2(a^2 + b^2) \]

### Step 8: Divide by \(a^2b^2c^2\)
Dividing through by \(a^2b^2c^2\) yields:
\[ \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2} \]

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

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