Find the equation to the circle
Whose radius is 3 and whose centre is ( - 1, 2).

Answer

To find the equation of the circle with a given center and radius, we can use the standard form of the equation of a circle, which is:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

where \((h, k)\) is the center of the circle and \(r\) is the radius.

### Step-by-Step Solution:

1. **Identify the center and radius**:
- The center of the circle is given as \((-1, 2)\), so \(h = -1\) and \(k = 2\).
- The radius \(r\) is given as \(3\).

2. **Substitute the values into the standard equation**:
- Substitute \(h\), \(k\), and \(r\) into the equation:
\[ (x - (-1))^2 + (y - 2)^2 = 3^2 \]
- This simplifies to:
\[ (x + 1)^2 + (y - 2)^2 = 9 \]

3. **Expand the equation**:
- Now, we will expand the left-hand side:
\[ (x + 1)^2 = x^2 + 2x + 1 \]
\[ (y - 2)^2 = y^2 - 4y + 4 \]
- Combining these, we have:
\[ x^2 + 2x + 1 + y^2 - 4y + 4 = 9 \]

4. **Combine like terms**:
- Combine all the terms on the left-hand side:
\[ x^2 + y^2 + 2x - 4y + 5 = 9 \]

5. **Set the equation to zero**:
- Move \(9\) to the left side:
\[ x^2 + y^2 + 2x - 4y + 5 - 9 = 0 \]
- This simplifies to:
\[ x^2 + y^2 + 2x - 4y - 4 = 0 \]

### Final Equation:
The equation of the circle is:
\[ x^2 + y^2 + 2x - 4y - 4 = 0 \]