A Delta is formed by the lines whose combined equation is given by `(x+y-9)(xy-2y-x+2)=0`
if the circumcentre of Delta is `(h,k)` then value of `|h-k|` is

Explanation

To solve the problem, we start with the combined equation of the lines:

\[ (x+y-9)(xy-2y-x+2) = 0 \]

This equation represents two lines. We can separate them as follows:

1. The first line is given by:
\[ x + y - 9 = 0 \quad \text{(Line 1)} \]

2. The second part can be simplified:
\[ xy - 2y - x + 2 = 0 \]
Rearranging gives:
\[ xy - x - 2y + 2 = 0 \implies y(x - 2) = x - 2 \]
This can be factored as:
\[ (y - 1)(x - 2) = 0 \]
Thus, we have:
\[ y - 1 = 0 \quad \text{(Line 2)} \quad \text{and} \quad x - 2 = 0 \quad \text{(Line 3)} \]

So, we have three lines:
- Line 1: \(x + y - 9 = 0\)
- Line 2: \(y = 1\)
- Line 3: \(x = 2\)

Next, we find the points of intersection of these lines to form the vertices of the triangle.

### Finding the Points of Intersection

1. **Point A (Intersection of Line 2 and Line 3)**:
- From Line 2: \(y = 1\)
- From Line 3: \(x = 2\)
- Thus, \(A(2, 1)\).

2. **Point B (Intersection of Line 1 and Line 2)**:
- Substitute \(y = 1\) into Line 1:
\[ x + 1 - 9 = 0 \implies x = 8 \]
- Thus, \(B(8, 1)\).

3. **Point C (Intersection of Line 1 and Line 3)**:
- Substitute \(x = 2\) into Line 1:
\[ 2 + y - 9 = 0 \implies y = 7 \]
- Thus, \(C(2, 7)\).

### Finding the Circumcenter

Since triangle ABC is a right triangle (as Line 2 and Line 3 are perpendicular), the circumcenter lies at the midpoint of the hypotenuse, which is segment BC.

1. **Coordinates of Points B and C**:
- \(B(8, 1)\)
- \(C(2, 7)\)

2. **Midpoint Formula**:
The midpoint \(M\) of segment \(BC\) is given by:
\[ M\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = M\left(\frac{8 + 2}{2}, \frac{1 + 7}{2}\right) \]
\[ M\left(\frac{10}{2}, \frac{8}{2}\right) = M(5, 4) \]

Thus, the circumcenter \( (h, k) = (5, 4) \).

### Finding \(|h - k|\)

Now, we calculate \(|h - k|\):
\[ |h - k| = |5 - 4| = |1| = 1 \]

### Final Answer
The value of \(|h - k|\) is \(1\).

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