The equation of the circle concentric with the circle `x^(2) + y^(2) - 6x - 4y - 12 =0` and touching y axis

Options

A. `x^(2) + y^(2) - 6x - 4y + 4 =0`
B. `x^(2) + y^(2) - 6x - 4y + 9 =0`
C. `x^(2) + y^(2) - 6x - 4y - 4 =0`
D. `x^(2) + y^(2) - 6x - 4y - 9 =0`

Explanation

To solve the problem of finding the equation of a circle that is concentric with the given circle and touches the y-axis, we can follow these steps:

### Step 1: Identify the center of the given circle
The equation of the given circle is:
\[ x^2 + y^2 - 6x - 4y - 12 = 0 \]

We can rewrite this equation in the standard form by completing the square.

### Step 2: Complete the square for x and y
1. For \( x^2 - 6x \):
- Take half of -6, which is -3, and square it to get 9.
- Thus, \( x^2 - 6x = (x - 3)^2 - 9 \).

2. For \( y^2 - 4y \):
- Take half of -4, which is -2, and square it to get 4.
- Thus, \( y^2 - 4y = (y - 2)^2 - 4 \).

Now substituting back into the equation:
\[ (x - 3)^2 - 9 + (y - 2)^2 - 4 - 12 = 0 \]
\[ (x - 3)^2 + (y - 2)^2 - 25 = 0 \]
\[ (x - 3)^2 + (y - 2)^2 = 25 \]

### Step 3: Identify the center and radius of the given circle
From the equation \((x - 3)^2 + (y - 2)^2 = 25\):
- The center \((h, k)\) is \((3, 2)\).
- The radius \(r_1\) is \(\sqrt{25} = 5\).

### Step 4: Determine the radius of the new circle
Since the new circle is concentric with the given circle, it will have the same center \((3, 2)\). The new circle touches the y-axis, which means the distance from the center to the y-axis must equal the radius of the new circle \(r_2\).

The distance from the center \((3, 2)\) to the y-axis (where \(x = 0\)) is:
\[ 3 - 0 = 3 \]
Thus, the radius \(r_2 = 3\).

### Step 5: Write the equation of the new circle
The equation of the new circle with center \((3, 2)\) and radius \(3\) is:
\[ (x - 3)^2 + (y - 2)^2 = 3^2 \]
\[ (x - 3)^2 + (y - 2)^2 = 9 \]

### Step 6: Expand the equation
Expanding the equation:
\[ (x - 3)^2 + (y - 2)^2 = 9 \]
\[ (x^2 - 6x + 9) + (y^2 - 4y + 4) = 9 \]
Combining terms:
\[ x^2 + y^2 - 6x - 4y + 13 - 9 = 0 \]
\[ x^2 + y^2 - 6x - 4y + 4 = 0 \]

### Final Answer
The equation of the circle is:
\[ x^2 + y^2 - 6x - 4y + 4 = 0 \]