A circle with center (2, 2) touches the coordinate axes and a straight line AB where A and B ie on direction of coordinate axes such that the lies between and the line AB be the origin then the locus of circumcenter of `triangle OAB` will be:

Options

A. `xy=x+y+sqrt(x^(2)+y^(2))`
B. `xy=x+y-sqrt(x^(2)+y^(2))`
C. `xy+x+y=sqrt(x^(2)+y^(2))`
D. `xy+x+y+sqrt(x^(2)+y^(2))=0`

Answer

To solve the problem, we need to find the locus of the circumcenter of triangle OAB, where O is the origin, A and B are points on the axes, and the circle with center (2, 2) touches the coordinate axes.

### Step-by-step Solution:

1. **Identify the Circle and its Properties**:
- The circle has a center at (2, 2) and touches the coordinate axes. Therefore, the radius of the circle is 2.
- The equation of the circle can be written as:
\[ (x - 2)^2 + (y - 2)^2 = 2^2 \]
- Simplifying gives:
\[ (x - 2)^2 + (y - 2)^2 = 4 \]

2. **Define Points A and B**:
- Let point A be (A, 0) on the x-axis and point B be (0, B) on the y-axis.
- The coordinates of points A and B are determined by the line AB.

3. **Find the Midpoint (Circumcenter)**:
- The circumcenter of triangle OAB is the midpoint of the line segment AB.
- The coordinates of the circumcenter (H, K) can be calculated as:
\[ H = \frac{A + 0}{2} = \frac{A}{2}, \quad K = \frac{0 + B}{2} = \frac{B}{2} \]
- Thus, the circumcenter is at:
\[ \left(\frac{A}{2}, \frac{B}{2}\right) \]

4. **Use the Condition of the Circle**:
- The circle touches the line AB, which means the perpendicular distance from the center of the circle (2, 2) to the line AB must equal the radius (2).
- The equation of line AB can be expressed as:
\[ y = -\frac{B}{A}x + B \]
- The general form of the line is:
\[ Ax + By - B = 0 \]

5. **Calculate the Distance from the Circle's Center to the Line**:
- The distance \(d\) from the point (2, 2) to the line \(Ax + By - B = 0\) is given by:
\[ d = \frac{|A(2) + B(2) - B|}{\sqrt{A^2 + B^2}} = \frac{|2A + 2B - B|}{\sqrt{A^2 + B^2}} = \frac{|2A + B|}{\sqrt{A^2 + B^2}} \]
- Setting this equal to the radius (2):
\[ \frac{|2A + B|}{\sqrt{A^2 + B^2}} = 2 \]

6. **Square Both Sides**:
- Squaring both sides gives:
\[ (2A + B)^2 = 4(A^2 + B^2) \]
- Expanding and rearranging leads to:
\[ 4A^2 + 4AB + B^2 = 4A^2 + 4B^2 \]
- Simplifying gives:
\[ 4AB + B^2 - 4B^2 = 0 \implies 4AB - 3B^2 = 0 \]
- Factoring out B:
\[ B(4A - 3B) = 0 \]
- This gives two cases: \(B = 0\) or \(4A - 3B = 0\).

7. **Find the Locus**:
- From \(4A - 3B = 0\), we can express \(B\) in terms of \(A\):
\[ B = \frac{4}{3}A \]
- Therefore, the locus of the circumcenter (H, K) where \(H = \frac{A}{2}\) and \(K = \frac{B}{2}\) becomes:
\[ K = \frac{4}{3} \cdot \frac{A}{2} = \frac{2}{3}A \]

8. **Final Equation**:
- Substituting \(A = 3K/2\) into \(H\) gives:
\[ H = \frac{3K}{4} \]
- Thus, the locus can be expressed as:
\[ 4H = 3K \quad \text{or} \quad 3K - 4H = 0 \]

### Conclusion:
The locus of the circumcenter of triangle OAB is given by the equation:
\[ 3K - 4H = 0 \]