A circle lies completely in the third quadrant and touches both the axes, its radius is given to be 7. Its centre is ..........

Answer

To find the center of a circle that lies completely in the third quadrant, touches both the axes, and has a radius of 7, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Position of the Circle**:
The circle is in the third quadrant, meaning both x and y coordinates of the center will be negative. Since the circle touches both the x-axis and y-axis, the distance from the center to each axis will be equal to the radius.

2. **Identify the Radius**:
The radius of the circle is given as \( R = 7 \).

3. **Determine the Center's Coordinates**:
- Since the circle touches the x-axis, the y-coordinate of the center will be equal to the negative radius: \( y = -R = -7 \).
- Similarly, since the circle touches the y-axis, the x-coordinate of the center will also be equal to the negative radius: \( x = -R = -7 \).
- Therefore, the coordinates of the center \( O \) are \( (-7, -7) \).

4. **Write the Equation of the Circle**:
The general equation of a circle with center \( (h, k) \) and radius \( r \) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \( h = -7 \), \( k = -7 \), and \( r = 7 \):
\[ (x + 7)^2 + (y + 7)^2 = 7^2 \]
Simplifying this gives:
\[ (x + 7)^2 + (y + 7)^2 = 49 \]

5. **Expand the Equation**:
Expanding the left side:
\[ (x^2 + 14x + 49) + (y^2 + 14y + 49) = 49 \]
Combining like terms:
\[ x^2 + y^2 + 14x + 14y + 98 = 49 \]
Rearranging gives:
\[ x^2 + y^2 + 14x + 14y + 49 = 0 \]

6. **Final Answer**:
The center of the circle is \( (-7, -7) \).