Find the centre and radius of each of the
circles whose equations are given below.
`x^(2) + y^(2) +2ax-2by + b^(2)=0`

Answer

To find the center and radius of the circle given by the equation \( x^2 + y^2 + 2ax - 2by + b^2 = 0 \), we can follow these steps:

### Step 1: Identify the general form of the circle's equation
The general equation of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \( (g, f) \) represents the center of the circle and \( c \) is a constant.

### Step 2: Compare the given equation with the general form
We have the equation:
\[ x^2 + y^2 + 2ax - 2by + b^2 = 0 \]
From this, we can identify:
- \( g = a \)
- \( f = -b \)
- \( c = b^2 \)

### Step 3: Find the center of the circle
The center of the circle \( (h, k) \) can be calculated using the formulas:
\[ h = -g = -a \]
\[ k = -f = b \]
Thus, the center of the circle is:
\[ \text{Center} = (-a, b) \]

### Step 4: Find the radius of the circle
The radius \( r \) can be calculated using the formula:
\[ r = \sqrt{g^2 + f^2 - c} \]
Substituting the values we found:
\[ r = \sqrt{a^2 + (-b)^2 - b^2} \]
This simplifies to:
\[ r = \sqrt{a^2 + b^2 - b^2} = \sqrt{a^2} = |a| \]
Since radius is a distance, we take the positive value:
\[ r = a \]

### Final Answer
Thus, the center and radius of the circle are:
- **Center**: \( (-a, b) \)
- **Radius**: \( a \)

---