The two circles `x^2 + y^2 -2x+6y+6=0 and x^2 + y^2 - 5x + 6y + 15 = 0` touch eachother. The equation of their common tangent is : (A) `x=3` (B) `y=6` (C) `7x-12y-21=0` (D) `7x+12y+21=0`

Options

A. x=3
B. y=6
C. `7x-12y-12=0`
D. `7x+12y+21=0`

Answer

To solve the problem of finding the common tangent to the two circles given by the equations \(x^2 + y^2 - 2x + 6y + 6 = 0\) and \(x^2 + y^2 - 5x + 6y + 15 = 0\), we will follow these steps:

### Step 1: Identify the equations of the circles
The equations of the circles are:
1. Circle 1: \(C_1: x^2 + y^2 - 2x + 6y + 6 = 0\)
2. Circle 2: \(C_2: x^2 + y^2 - 5x + 6y + 15 = 0\)

### Step 2: Rewrite the equations in standard form
To find the centers and radii of the circles, we will rewrite the equations in the standard form \((x - h)^2 + (y - k)^2 = r^2\).

#### For Circle 1:
Starting with \(C_1\):
\[ x^2 - 2x + y^2 + 6y + 6 = 0 \]
Completing the square for \(x\) and \(y\):
\[ (x^2 - 2x + 1) + (y^2 + 6y + 9) = -6 + 1 + 9 \]
\[ (x - 1)^2 + (y + 3)^2 = 4 \]
Thus, the center \(C_1\) is \((1, -3)\) and the radius \(r_1 = 2\).

#### For Circle 2:
Starting with \(C_2\):
\[ x^2 - 5x + y^2 + 6y + 15 = 0 \]
Completing the square for \(x\) and \(y\):
\[ (x^2 - 5x + \frac{25}{4}) + (y^2 + 6y + 9) = -15 + \frac{25}{4} + 9 \]
\[ (x - \frac{5}{2})^2 + (y + 3)^2 = \frac{1}{4} \]
Thus, the center \(C_2\) is \((\frac{5}{2}, -3)\) and the radius \(r_2 = \frac{1}{2}\).

### Step 3: Calculate the distance between the centers
The distance \(d\) between the centers \(C_1(1, -3)\) and \(C_2(\frac{5}{2}, -3)\) is calculated as follows:
\[ d = \sqrt{\left(\frac{5}{2} - 1\right)^2 + (-3 - (-3))^2} = \sqrt{\left(\frac{3}{2}\right)^2} = \frac{3}{2} \]

### Step 4: Check the condition for tangency
For the circles to touch externally, the distance between the centers must equal the sum of the radii:
\[ d = r_1 + r_2 \implies \frac{3}{2} = 2 + \frac{1}{2} = \frac{5}{2} \]
This condition does not hold, indicating that the circles touch internally.

### Step 5: Find the equation of the common tangent
The common tangent can be found using the radical axis method. The radical axis of two circles can be found by equating their equations.

Setting \(C_1 = C_2\):
\[ x^2 + y^2 - 2x + 6y + 6 = x^2 + y^2 - 5x + 6y + 15 \]
Cancelling \(x^2\) and \(y^2\) and simplifying:
\[ -2x + 6 = -5x + 15 \]
Rearranging gives:
\[ 5x - 2x = 15 - 6 \implies 3x = 9 \implies x = 3 \]

### Conclusion
The equation of the common tangent is:
\[ x = 3 \]

### Final Answer
Thus, the correct option is (A) \(x = 3\).