find the equation of circle in each of the following cases
(b) touches both the coordinate axes in second quadrant and having radius 2

Answer

To find the equation of a circle that touches both coordinate axes in the second quadrant and has a radius of 2, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Position of the Circle**:
Since the circle touches both the x-axis and y-axis in the second quadrant, the center of the circle must be located at a point where both coordinates are negative. The radius of the circle is given as 2.

2. **Determine the Center of the Circle**:
The distance from the center of the circle to the x-axis (which is the y-coordinate of the center) must be equal to the radius. Similarly, the distance from the center to the y-axis (which is the x-coordinate of the center) must also equal the radius. Therefore, if the center is at point (h, k), we have:
- k = 2 (distance to the x-axis)
- h = -2 (distance to the y-axis)

Hence, the coordinates of the center of the circle are:
\[ (h, k) = (-2, 2) \]

3. **Use the Circle Equation**:
The standard form of the equation of a circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where (h, k) is the center and r is the radius. Substituting the values we found:
- h = -2
- k = 2
- r = 2

The equation becomes:
\[ (x - (-2))^2 + (y - 2)^2 = 2^2 \]
Simplifying this, we get:
\[ (x + 2)^2 + (y - 2)^2 = 4 \]

4. **Final Equation**:
Thus, the equation of the circle that touches both coordinate axes in the second quadrant and has a radius of 2 is:
\[ (x + 2)^2 + (y - 2)^2 = 4 \]