The centres of the three circles `x^2 + y^2 - 10x + 9 = 0, x² + y^2 - 6x + 2y + 1 = 0, x^2 + y^2 - 9x - 4y +2=0`

Options

A. lie on the straight line `x-2y=5`
B. lie on circle `x^(2)+y^(2)=25`
C. do not lie on straight line
D. lie on circle `x^(2)+y^(2)+x+y-17=0`

Answer

To find the centers of the three circles given by their equations, we will follow these steps:

### Step 1: Rewrite the equations of the circles in standard form
The general equation of a circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center and \(r\) is the radius.

1. **Circle 1**: \(x^2 + y^2 - 10x + 9 = 0\)
- Rearranging gives:
\[ x^2 - 10x + y^2 + 9 = 0 \]
- Completing the square for \(x\):
\[ (x^2 - 10x + 25) + y^2 = 16 \]
- Thus, the center is \((5, 0)\).

2. **Circle 2**: \(x^2 + y^2 - 6x + 2y + 1 = 0\)
- Rearranging gives:
\[ x^2 - 6x + y^2 + 2y + 1 = 0 \]
- Completing the square:
\[ (x^2 - 6x + 9) + (y^2 + 2y + 1) = 9 \]
- Thus, the center is \((3, -1)\).

3. **Circle 3**: \(x^2 + y^2 - 9x - 4y + 2 = 0\)
- Rearranging gives:
\[ x^2 - 9x + y^2 - 4y + 2 = 0 \]
- Completing the square:
\[ (x^2 - 9x + 20) + (y^2 - 4y + 4) = 22 \]
- Thus, the center is \((\frac{18}{2}, 2)\) or \((9, 2)\).

### Step 2: Identify the centers
The centers of the circles are:
- Circle 1: \(C_1(5, 0)\)
- Circle 2: \(C_2(3, -1)\)
- Circle 3: \(C_3(9, 2)\)

### Step 3: Check if the centers are collinear
To check if the points \(C_1(5, 0)\), \(C_2(3, -1)\), and \(C_3(9, 2)\) are collinear, we can use the area of the triangle formed by these points. The area can be calculated using the determinant formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates:
\[ \text{Area} = \frac{1}{2} \left| 5(-1 - 2) + 3(2 - 0) + 9(0 + 1) \right| \]
Calculating:
\[ = \frac{1}{2} \left| 5(-3) + 3(2) + 9(1) \right| = \frac{1}{2} \left| -15 + 6 + 9 \right| = \frac{1}{2} \left| 0 \right| = 0 \]
Since the area is zero, the points are collinear.

### Conclusion
The centers of the three circles are collinear.