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

Answer

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
The equation of the given circle is:
\[ x^2 + y^2 - 6x - 6y + 14 = 0 \]
We can rewrite it in standard form by completing the square.

### Step 2: Complete the square
1. Rearranging the equation:
\[ (x^2 - 6x) + (y^2 - 6y) + 14 = 0 \]
2. Completing the square for \(x\) and \(y\):
\[ (x - 3)^2 - 9 + (y - 3)^2 - 9 + 14 = 0 \]
\[ (x - 3)^2 + (y - 3)^2 - 4 = 0 \]
\[ (x - 3)^2 + (y - 3)^2 = 4 \]

### Step 3: Determine the center and radius of the given circle
From the standard form \((x - 3)^2 + (y - 3)^2 = 4\):
- The center \(C_1\) of the given circle is \((3, 3)\).
- The radius \(r_1\) is \(\sqrt{4} = 2\).

### Step 4: Set up the conditions for the new circle
Let the center of the new circle be \((h, k)\) and its radius be \(r_2\). Since the new circle touches the y-axis, the radius \(r_2\) is equal to the x-coordinate of the center:
\[ r_2 = h \]

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

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

### Step 7: Expand both sides
1. Left side:
\[ (h - 3)^2 + (k - 3)^2 = h^2 - 6h + 9 + k^2 - 6k + 9 \]
\[ = h^2 + k^2 - 6h - 6k + 18 \]
2. Right side:
\[ (2 + h)^2 = 4 + 4h + h^2 \]

### Step 8: Set the two sides equal
Equating both sides:
\[ h^2 + k^2 - 6h - 6k + 18 = 4 + 4h + h^2 \]

### Step 9: Simplify the equation
1. Cancel \(h^2\) from both sides:
\[ k^2 - 6h - 6k + 18 = 4 + 4h \]
2. Rearranging gives:
\[ k^2 - 6k - 8 - 8h = 0 \]

### Step 10: Replace \(h\) and \(k\) with \(x\) and \(y\)
Substituting \(h\) with \(x\) and \(k\) with \(y\) gives:
\[ y^2 - 6y - 8 - 8x = 0 \]
Rearranging gives:
\[ y^2 - 6y - 8x + 8 = 0 \]

### Final Step: Rearranging to standard form
Rearranging gives us the locus of the center:
\[ x + \frac{y^2 - 6y + 8}{8} = 0 \]