Locus of the centre of the circle which touches `x^2+y^2 - 6x-6y+14 =0` externally and also y-axis is:

Options

A. `x^(2)-6x-10y-14=0`
B. `x^(2)-10x-6y-14=0`
C. `y^(2)-6x-10y+14=0`
D. `y^(2)-10x-6y+14=0`

Explanation

To find the locus of the center of a circle that touches the given circle externally and also touches the y-axis, we can follow these steps:

### Step 1: Identify the given circle's equation
The equation of the given circle is:
\[ x^2 + y^2 - 6x - 6y + 14 = 0 \]

### Step 2: Rewrite the equation in standard form
To rewrite the equation in standard form, we will complete the square for both \(x\) and \(y\).

1. Group the \(x\) terms and the \(y\) terms:
\[ (x^2 - 6x) + (y^2 - 6y) + 14 = 0 \]

2. Complete the square:
- For \(x^2 - 6x\):
\[ x^2 - 6x = (x - 3)^2 - 9 \]
- For \(y^2 - 6y\):
\[ y^2 - 6y = (y - 3)^2 - 9 \]

3. Substitute back into the equation:
\[ (x - 3)^2 - 9 + (y - 3)^2 - 9 + 14 = 0 \]
\[ (x - 3)^2 + (y - 3)^2 - 4 = 0 \]
\[ (x - 3)^2 + (y - 3)^2 = 4 \]

### Step 3: Identify the center and radius of the given circle
From the standard form \((x - 3)^2 + (y - 3)^2 = 4\):
- The center \(C_1\) is at \((3, 3)\).
- The radius \(r_1\) is \(2\) (since \(r^2 = 4\)).

### Step 4: Set up the locus of the center of the new circle
Let the center of the new circle be \((h, k)\). Since the new circle touches the y-axis, the distance from the center to the y-axis must equal the radius \(r_2\) of the new circle:
\[ h = r_2 \]

### Step 5: Use the condition for external tangency
The distance \(d\) between the centers \(C_1(3, 3)\) and \(C_2(h, k)\) must equal the sum of the radii:
\[ d = r_1 + r_2 \]
\[ \sqrt{(3 - h)^2 + (3 - k)^2} = 2 + r_2 \]

Substituting \(r_2 = h\) into the equation gives:
\[ \sqrt{(3 - h)^2 + (3 - k)^2} = 2 + h \]

### Step 6: Square both sides to eliminate the square root
Squaring both sides:
\[ (3 - h)^2 + (3 - k)^2 = (2 + h)^2 \]

### Step 7: Expand both sides
1. Left side:
\[ (3 - h)^2 + (3 - k)^2 = (3 - h)^2 + (3 - k)^2 \]
\[ = 9 - 6h + h^2 + 9 - 6k + k^2 = h^2 + k^2 - 6h - 6k + 18 \]

2. Right side:
\[ (2 + h)^2 = 4 + 4h + h^2 \]

### Step 8: Set the equations equal and simplify
Setting both sides equal:
\[ h^2 + k^2 - 6h - 6k + 18 = 4 + 4h + h^2 \]
Cancel \(h^2\) from both sides:
\[ k^2 - 6k + 18 = 4 + 4h \]
Rearranging gives:
\[ k^2 - 6k - 4h + 14 = 0 \]

### Step 9: Express the locus in terms of \(x\) and \(y\)
Replace \(h\) with \(x\) and \(k\) with \(y\):
\[ y^2 - 6y - 4x + 14 = 0 \]

### Final Answer
The locus of the center of the circle is:
\[ y^2 - 6y - 4x + 14 = 0 \]