A triangle are` (6,0).(0, 6)` and `(6,6)`. If distance between circumcentre and orthocenter and distance between circumcentre and centroid are `lambda`and `u` unit respectively, then `(lambda, u)` lies on:

Options

A. `2sqrt(2)`
B. `2`
C. `3sqrt(3)`
D. 1

Explanation

To solve the problem, we need to find the distances between the circumcenter, orthocenter, and centroid of the triangle formed by the points (6,0), (0,6), and (6,6).

### Step-by-Step Solution:

1. **Identify the vertices of the triangle**:
The vertices of the triangle are given as:
- A(6, 0)
- B(0, 6)
- C(6, 6)

2. **Determine the type of triangle**:
By observing the coordinates, we can see that this triangle is a right triangle with the right angle at C(6, 6).

3. **Find the Orthocenter (H)**:
For a right triangle, the orthocenter is located at the vertex where the right angle is. Thus, the orthocenter H is at point C:
- H = (6, 6)

4. **Find the Circumcenter (O)**:
The circumcenter of a right triangle is located at the midpoint of the hypotenuse. The hypotenuse is AB, which connects points A(6, 0) and B(0, 6).
- Midpoint O = \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\)
- O = \(\left(\frac{6 + 0}{2}, \frac{0 + 6}{2}\right) = (3, 3)\)

5. **Find the Centroid (G)**:
The centroid of a triangle is given by the formula:
- G = \(\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)\)
- G = \(\left(\frac{6 + 0 + 6}{3}, \frac{0 + 6 + 6}{3}\right) = \left(\frac{12}{3}, \frac{12}{3}\right) = (4, 4)\)

6. **Calculate the distance between Circumcenter (O) and Orthocenter (H)**:
Using the distance formula:
- Distance \(d_{OH} = \sqrt{(x_H - x_O)^2 + (y_H - y_O)^2}\)
- \(d_{OH} = \sqrt{(6 - 3)^2 + (6 - 3)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}\)
- Let this distance be \(\lambda = 3\sqrt{2}\).

7. **Calculate the distance between Circumcenter (O) and Centroid (G)**:
Using the distance formula again:
- Distance \(d_{OG} = \sqrt{(x_G - x_O)^2 + (y_G - y_O)^2}\)
- \(d_{OG} = \sqrt{(4 - 3)^2 + (4 - 3)^2} = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}\)
- Let this distance be \(u = \sqrt{2}\).

8. **Conclusion**:
The distances \(\lambda\) and \(u\) are:
- \(\lambda = 3\sqrt{2}\)
- \(u = \sqrt{2}\)

Thus, the pair \((\lambda, u) = (3\sqrt{2}, \sqrt{2})\).

### Final Answer:
The pair \((\lambda, u)\) lies on the line defined by the relationship between these two distances.