A circle touches the line `y = x` at the point `(2, 2)` and has it centre on y-axis, then square of its radius is

Explanation

To solve the problem, we need to find the square of the radius of a circle that touches the line \(y = x\) at the point \((2, 2)\) and has its center on the y-axis.

### Step-by-Step Solution:

1. **Identify the center of the circle**:
Since the center of the circle lies on the y-axis, we can denote the center as \((0, k)\) where \(k\) is some value on the y-axis.

2. **Use the point of tangency**:
The circle touches the line \(y = x\) at the point \((2, 2)\). This means that the distance from the center of the circle to the line \(y = x\) is equal to the radius \(r\).

3. **Calculate the distance from the center to the line**:
The formula for the distance \(d\) from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For the line \(y = x\), we can rewrite it as \(x - y = 0\), which gives us \(A = 1\), \(B = -1\), and \(C = 0\). The center of the circle is \((0, k)\). Thus, the distance from the center to the line is:
\[ d = \frac{|1 \cdot 0 + (-1) \cdot k + 0|}{\sqrt{1^2 + (-1)^2}} = \frac{| -k |}{\sqrt{2}} = \frac{|k|}{\sqrt{2}} \]

4. **Set the distance equal to the radius**:
The radius \(r\) is also the distance from the center \((0, k)\) to the point of tangency \((2, 2)\). Using the distance formula:
\[ r = \sqrt{(2 - 0)^2 + (2 - k)^2} = \sqrt{4 + (2 - k)^2} \]

5. **Equate the two expressions for the radius**:
We have:
\[ r = \frac{|k|}{\sqrt{2}} \quad \text{and} \quad r = \sqrt{4 + (2 - k)^2} \]
Setting these equal gives:
\[ \frac{|k|}{\sqrt{2}} = \sqrt{4 + (2 - k)^2} \]

6. **Square both sides to eliminate the square root**:
\[ \frac{k^2}{2} = 4 + (2 - k)^2 \]
Expanding the right side:
\[ \frac{k^2}{2} = 4 + (4 - 4k + k^2) \]
Simplifying gives:
\[ \frac{k^2}{2} = 8 - 4k + k^2 \]
Multiplying through by 2 to eliminate the fraction:
\[ k^2 = 16 - 8k + 2k^2 \]
Rearranging gives:
\[ 0 = k^2 - 8k + 16 \]

7. **Solve the quadratic equation**:
Using the quadratic formula \(k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ k = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1} = \frac{8 \pm \sqrt{64 - 64}}{2} = \frac{8 \pm 0}{2} = 4 \]
So, \(k = 4\).

8. **Find the radius**:
Now, substituting \(k = 4\) back into the expression for the radius:
\[ r = \sqrt{(2 - 0)^2 + (2 - 4)^2} = \sqrt{4 + 4} = \sqrt{8} \]

9. **Calculate the square of the radius**:
The square of the radius is:
\[ r^2 = 8 \]

### Final Answer:
The square of the radius of the circle is \(8\).