The centres of the circles `x ^(2) + y ^(2) =1, x ^(2) + y^(2)+ 6x - 2y =1 and x ^(2) + y^(2) -12 x + 4y =1` are

Options

A. Same
B. Collinear
C. Non-collinear
D. None of these

Answer

To find the centers of the given circles, we will first rewrite each circle's equation in standard form. The standard form of a circle's equation is given by:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

where \((h, k)\) is the center of the circle and \(r\) is the radius.

### Step 1: Identify the center of the first circle

The first circle's equation is:

\[ x^2 + y^2 = 1 \]

This can be rewritten as:

\[ (x - 0)^2 + (y - 0)^2 = 1^2 \]

From this, we can see that the center \((h, k)\) is:

\[ (0, 0) \]

### Step 2: Identify the center of the second circle

The second circle's equation is:

\[ x^2 + y^2 + 6x - 2y = 1 \]

We will rearrange this equation to complete the square:

1. Group the \(x\) terms and the \(y\) terms:

\[ (x^2 + 6x) + (y^2 - 2y) = 1 \]

2. Complete the square for \(x\):

\[ x^2 + 6x = (x + 3)^2 - 9 \]

3. Complete the square for \(y\):

\[ y^2 - 2y = (y - 1)^2 - 1 \]

4. Substitute back into the equation:

\[ ((x + 3)^2 - 9) + ((y - 1)^2 - 1) = 1 \]

5. Simplify:

\[ (x + 3)^2 + (y - 1)^2 - 10 = 1 \]

\[ (x + 3)^2 + (y - 1)^2 = 11 \]

From this, the center is:

\[ (-3, 1) \]

### Step 3: Identify the center of the third circle

The third circle's equation is:

\[ x^2 + y^2 - 12x + 4y = 1 \]

Again, we will rearrange this equation to complete the square:

1. Group the \(x\) terms and the \(y\) terms:

\[ (x^2 - 12x) + (y^2 + 4y) = 1 \]

2. Complete the square for \(x\):

\[ x^2 - 12x = (x - 6)^2 - 36 \]

3. Complete the square for \(y\):

\[ y^2 + 4y = (y + 2)^2 - 4 \]

4. Substitute back into the equation:

\[ ((x - 6)^2 - 36) + ((y + 2)^2 - 4) = 1 \]

5. Simplify:

\[ (x - 6)^2 + (y + 2)^2 - 40 = 1 \]

\[ (x - 6)^2 + (y + 2)^2 = 41 \]

From this, the center is:

\[ (6, -2) \]

### Final Centers

Thus, the centers of the circles are:

1. Circle 1: \((0, 0)\)
2. Circle 2: \((-3, 1)\)
3. Circle 3: \((6, -2)\)