AB is any chord of the circle `x^2+y^2-6x-8y-11=0` which subtends an angle `pi/2` at `(1,2)`. If locus of midpoint of AB is a circle `x^2+y^2-2ax-2by-c=0`; then find the value of `(a+b+c)`.

Answer

To solve the problem, we need to find the value of \( (a + b + c) \) given the equation of the circle and the conditions provided.

### Step 1: Identify the center and radius of the circle
The equation of the circle is given as:
\[ x^2 + y^2 - 6x - 8y - 11 = 0 \]
We can rewrite this in standard form by completing the square.

1. Rearranging the equation:
\[ x^2 - 6x + y^2 - 8y = 11 \]

2. Completing the square for \(x\):
\[ x^2 - 6x = (x - 3)^2 - 9 \]

3. Completing the square for \(y\):
\[ y^2 - 8y = (y - 4)^2 - 16 \]

4. Substituting back:
\[ (x - 3)^2 - 9 + (y - 4)^2 - 16 = 11 \]
\[ (x - 3)^2 + (y - 4)^2 = 36 \]

From this, we see that the center of the circle is \( (3, 4) \) and the radius is \( 6 \).

### Step 2: Set up the condition for the chord
Let \( A \) and \( B \) be points on the circle such that the chord \( AB \) subtends a right angle at the point \( (1, 2) \). The midpoint \( M(h, k) \) of the chord \( AB \) will lie on the perpendicular bisector of \( AB \).

### Step 3: Use the right angle condition
Since \( AB \) subtends a right angle at \( (1, 2) \), we can use the property that the angle subtended by a chord at any point on the circle is half the angle subtended at the center. Therefore, the distance from the center \( (3, 4) \) to the point \( (1, 2) \) can be calculated:
\[ OM = \sqrt{(3 - 1)^2 + (4 - 2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]

### Step 4: Apply Pythagorean theorem
Using the Pythagorean theorem:
\[ OA^2 = OM^2 + AM^2 \]
Where \( OA = 6 \) (radius), \( OM = d \) (distance from center to midpoint), and \( AM \) is the distance from the midpoint to the point on the circle.

Thus:
\[ 6^2 = (2\sqrt{2})^2 + AM^2 \]
\[ 36 = 8 + AM^2 \]
\[ AM^2 = 36 - 8 = 28 \]

### Step 5: Find the locus of midpoint
The locus of the midpoint \( M(h, k) \) can be expressed as:
\[ AM^2 = (h - 1)^2 + (k - 2)^2 \]
Setting \( AM^2 = 28 \):
\[ (h - 1)^2 + (k - 2)^2 = 28 \]

### Step 6: Rewrite in standard form
This represents a circle centered at \( (1, 2) \) with radius \( \sqrt{28} \):
\[ (h - 1)^2 + (k - 2)^2 = 28 \]
Expanding this gives:
\[ h^2 - 2h + 1 + k^2 - 4k + 4 = 28 \]
\[ h^2 + k^2 - 2h - 4k - 23 = 0 \]

### Step 7: Compare with the given locus equation
The locus is given as:
\[ x^2 + y^2 - 2ax - 2by - c = 0 \]
From our equation, we can identify:
\[ -2a = -2 \implies a = 1 \]
\[ -2b = -4 \implies b = 2 \]
\[ -c = -23 \implies c = 23 \]

### Step 8: Calculate \( a + b + c \)
Now we can find \( a + b + c \):
\[ a + b + c = 1 + 2 + 23 = 26 \]

Thus, the final answer is:
\[ \boxed{26} \]