Find the equation to the circle
Whose radius is 10 and whose centre is ( - 5, - 6).

Answer

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

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

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 \((-5, -6)\).
- The radius of the circle is given as \(10\).

2. **Substitute the values into the standard equation**:
- Here, \(h = -5\), \(k = -6\), and \(r = 10\).
- Substitute these values into the standard form:
\[ (x - (-5))^2 + (y - (-6))^2 = 10^2 \]
- This simplifies to:
\[ (x + 5)^2 + (y + 6)^2 = 100 \]

3. **Expand the equation**:
- Now we will expand the left-hand side:
\[ (x + 5)^2 = x^2 + 10x + 25 \]
\[ (y + 6)^2 = y^2 + 12y + 36 \]
- Combining these gives:
\[ x^2 + 10x + 25 + y^2 + 12y + 36 = 100 \]

4. **Combine like terms**:
- Combine the constant terms on the left:
\[ x^2 + y^2 + 10x + 12y + (25 + 36) = 100 \]
\[ x^2 + y^2 + 10x + 12y + 61 = 100 \]

5. **Rearrange the equation**:
- Move \(100\) to the left side:
\[ x^2 + y^2 + 10x + 12y + 61 - 100 = 0 \]
\[ x^2 + y^2 + 10x + 12y - 39 = 0 \]

### Final Equation:
The equation of the circle is:
\[ x^2 + y^2 + 10x + 12y - 39 = 0 \]