Find the equation of the circle having centre in the first quadrant, touching the x-axis, having a common tangent `y = sqrt3 x + 4` with the circle `x ^(2) + y ^(2) + 4x + 4y + 4=0` such that the distance between the two circles along the x-axis is 3 units.

Answer

To find the equation of the circle with the given conditions, we will follow these steps:

### Step 1: Identify the center of the first circle
The first circle is given by the equation:
\[ x^2 + y^2 + 4x + 4y + 4 = 0 \]
We can rewrite this in standard form. Completing the square for \(x\) and \(y\):
\[ (x^2 + 4x) + (y^2 + 4y) + 4 = 0 \]
\[ (x + 2)^2 - 4 + (y + 2)^2 - 4 + 4 = 0 \]
\[ (x + 2)^2 + (y + 2)^2 = 4 \]
This shows that the center of the first circle is at \((-2, -2)\) and the radius \(r_1 = 2\).

### Step 2: Determine the center and radius of the second circle
Let the center of the second circle be \((h, k)\) and its radius \(r_2 = k\) (since it touches the x-axis).

### Step 3: Use the common tangent condition
The line \(y = \sqrt{3}x + 4\) can be rewritten in standard form:
\[ \sqrt{3}x - y + 4 = 0 \]
The distance from the center \((h, k)\) of the second circle to this line must equal the radius \(k\):
\[ \text{Distance} = \frac{|\sqrt{3}h - k + 4|}{\sqrt{(\sqrt{3})^2 + (-1)^2}} = \frac{|\sqrt{3}h - k + 4|}{2} \]
Setting this equal to \(k\):
\[ \frac{|\sqrt{3}h - k + 4|}{2} = k \]
Multiplying through by 2:
\[ |\sqrt{3}h - k + 4| = 2k \]
This gives us two cases to consider:
1. \(\sqrt{3}h - k + 4 = 2k\)
2. \(\sqrt{3}h - k + 4 = -2k\)

### Step 4: Solve the first case
From the first case:
\[ \sqrt{3}h + 4 = 3k \implies k = \frac{\sqrt{3}h + 4}{3} \]

### Step 5: Use the distance condition along the x-axis
The distance between the two circles along the x-axis is given as 3 units. The distance between the centers is:
\[ |h + 2| = r_1 + r_2 + 3 = 2 + k + 3 = k + 5 \]
Thus:
\[ |h + 2| = k + 5 \]

### Step 6: Substitute for \(k\)
Substituting \(k = \frac{\sqrt{3}h + 4}{3}\) into the distance equation:
\[ |h + 2| = \frac{\sqrt{3}h + 4}{3} + 5 \]
Multiply through by 3 to eliminate the fraction:
\[ 3|h + 2| = \sqrt{3}h + 4 + 15 \]
\[ 3|h + 2| = \sqrt{3}h + 19 \]

### Step 7: Solve the absolute value equation
This gives us two cases to solve:
1. \(3(h + 2) = \sqrt{3}h + 19\)
2. \(3(-h - 2) = \sqrt{3}h + 19\)

### Step 8: Solve the first case
From the first case:
\[ 3h + 6 = \sqrt{3}h + 19 \]
\[ (3 - \sqrt{3})h = 13 \implies h = \frac{13}{3 - \sqrt{3}} \]

### Step 9: Solve the second case
From the second case:
\[ -3h - 6 = \sqrt{3}h + 19 \]
\[ -3h - \sqrt{3}h = 25 \implies (-3 - \sqrt{3})h = 25 \implies h = \frac{-25}{3 + \sqrt{3}} \]

### Step 10: Calculate \(k\) for both cases
Substituting \(h\) back into \(k = \frac{\sqrt{3}h + 4}{3}\) for both values of \(h\) to find \(k\).

### Step 11: Write the equation of the circle
The equation of the circle is given by:
\[ (x - h)^2 + (y - k)^2 = k^2 \]

### Final Equation
After calculating \(h\) and \(k\), substitute back to get the final equation of the circle.