If the circles `(x-a)^(2)+(y-b)^(2)=c^(2)` and `(x-b)^(2)+(y-a)^(2)=c^(2)` touch each other, then

Options

A. `a=bpm 2c`
B. `a=b pm sqrt(2)c`
C. `a=b pm c`
D. none of these

Answer

To solve the problem, we need to analyze the two given circles and determine the condition under which they touch each other.

### Step 1: Identify the centers and radii of the circles
The equations of the circles are:
1. \((x - a)^2 + (y - b)^2 = c^2\)
2. \((x - b)^2 + (y - a)^2 = c^2\)

From the first equation, we can identify:
- Center of Circle 1: \(C_1(a, b)\)
- Radius of Circle 1: \(r_1 = c\)

From the second equation, we can identify:
- Center of Circle 2: \(C_2(b, a)\)
- Radius of Circle 2: \(r_2 = c\)

### Step 2: Calculate the distance between the centers
The distance \(d\) between the centers \(C_1(a, b)\) and \(C_2(b, a)\) can be calculated using the distance formula:
\[ d = \sqrt{(b - a)^2 + (a - b)^2} \]
This simplifies to:
\[ d = \sqrt{(b - a)^2 + (b - a)^2} = \sqrt{2(b - a)^2} = |b - a| \sqrt{2} \]

### Step 3: Set up the condition for the circles to touch
For two circles to touch each other externally, the distance between their centers must equal the sum of their radii. Since both circles have the same radius \(c\), the condition becomes:
\[ d = r_1 + r_2 = c + c = 2c \]
Thus, we have:
\[ |b - a| \sqrt{2} = 2c \]

### Step 4: Solve for \(a\) in terms of \(b\) and \(c\)
Squaring both sides gives:
\[ (b - a)^2 \cdot 2 = (2c)^2 \]
This simplifies to:
\[ 2(b - a)^2 = 4c^2 \]
Dividing both sides by 2:
\[ (b - a)^2 = 2c^2 \]
Taking the square root of both sides:
\[ |b - a| = \sqrt{2}c \]
This leads to two equations:
1. \(b - a = \sqrt{2}c\)
2. \(b - a = -\sqrt{2}c\)

From these, we can express \(a\) in terms of \(b\) and \(c\):
1. \(a = b - \sqrt{2}c\)
2. \(a = b + \sqrt{2}c\)

### Conclusion
Thus, the values of \(a\) can be expressed as:
\[ a = b \pm \sqrt{2}c \]