Through the point P(3,4) a pair of perpendicular lines are drawn which meet x-axis at the point A and B. The locus of incentre of triangle PAB is (a) `x^(2)-y^(2)-6x-8y+25=0` (b) `x^(2)+y^(2)-6x-8y+25=0` (c) `x^(2)-y^(2)+6x+8y+25=0` (d) `x^(2)+y^(2)+6x+8y+25=0`

Options

A. `x^(2)-y^(2)-6x-8y+25=0`
B. `x^(2)+y^(2)-6x-8y+25=0`
C. `x^(2)-y^(2)+6x+8y+25=0`
D. `x^(2)+y^(2)+6x+8y+25=0`

Explanation

To find the locus of the incentre of triangle PAB, we will follow these steps:

### Step 1: Understand the Geometry
We have a point P(3, 4) and two perpendicular lines drawn from this point, meeting the x-axis at points A and B. The incentre I of triangle PAB is the point where the angle bisectors of the triangle intersect.

### Step 2: Define the Equations of the Lines
Let the slopes of the two perpendicular lines be m and -1/m. The equations of the lines passing through point P(3, 4) can be expressed as:
1. Line 1: \( y - 4 = m(x - 3) \)
2. Line 2: \( y - 4 = -\frac{1}{m}(x - 3) \)

### Step 3: Find Points A and B
To find where these lines intersect the x-axis (where y = 0), we set y = 0 in the equations of the lines:
1. For Line 1:
\[ 0 - 4 = m(x - 3) \implies x = \frac{4}{m} + 3 \]
Let this point be A.

2. For Line 2:
\[ 0 - 4 = -\frac{1}{m}(x - 3) \implies x = -4m + 3 \]
Let this point be B.

### Step 4: Coordinates of Points A and B
Thus, the coordinates of points A and B are:
- A: \( \left(\frac{4}{m} + 3, 0\right) \)
- B: \( \left(-4m + 3, 0\right) \)

### Step 5: Find the Incentre I
The coordinates of the incentre I of triangle PAB can be found using the formula:
\[ I = \left(\frac{aA_x + bB_x + cP_x}{a + b + c}, \frac{aA_y + bB_y + cP_y}{a + b + c}\right) \]
where a, b, and c are the lengths of sides opposite to vertices A, B, and P respectively.

### Step 6: Calculate the Lengths of the Sides
Using the distance formula:
- Length PA:
\[ PA = \sqrt{\left(\frac{4}{m} + 3 - 3\right)^2 + (0 - 4)^2} = \sqrt{\left(\frac{4}{m}\right)^2 + 16} \]
- Length PB:
\[ PB = \sqrt{\left(-4m + 3 - 3\right)^2 + (0 - 4)^2} = \sqrt{(-4m)^2 + 16} = \sqrt{16m^2 + 16} \]
- Length AB:
\[ AB = \sqrt{\left(\left(\frac{4}{m} + 3\right) - (-4m + 3)\right)^2 + (0 - 0)^2} = \sqrt{\left(\frac{4}{m} + 4m\right)^2} \]

### Step 7: Substitute into the Incentre Formula
Substituting the coordinates of A, B, and P along with the lengths calculated into the incentre formula will yield the coordinates of I in terms of m.

### Step 8: Eliminate Parameter m
To find the locus, we eliminate the parameter m. This will involve substituting the expressions for the coordinates of I back into the equations and simplifying.

### Step 9: Final Equation
After simplification, we will arrive at the equation of the locus of the incentre I. The final equation will be:
\[ x^2 - y^2 - 6x - 8y + 25 = 0 \]

### Conclusion
Thus, the locus of the incentre of triangle PAB is given by option (a):
\[ \text{(a) } x^2 - y^2 - 6x - 8y + 25 = 0 \]