How many common tangents can be drawn to the following circles `x^(2)+y^(2)=6x` and `x^(2)+y^(2)+6x+2y+1=0`?

Options

A. 4
B. 3
C. 2
D. 1

Explanation

To find the number of common tangents that can be drawn to the circles given by the equations \(x^2 + y^2 = 6x\) and \(x^2 + y^2 + 6x + 2y + 1 = 0\), we will follow these steps:

### Step 1: Rewrite the equations in standard form
1. **For the first circle**:
\[ x^2 + y^2 - 6x = 0 \implies (x^2 - 6x + 9) + y^2 = 9 \implies (x - 3)^2 + y^2 = 3^2 \]
- Center \(C_1 = (3, 0)\)
- Radius \(r_1 = 3\)

2. **For the second circle**:
\[ x^2 + y^2 + 6x + 2y + 1 = 0 \implies (x^2 + 6x + 9) + (y^2 + 2y + 1) = 9 \implies (x + 3)^2 + (y + 1)^2 = 3^2 \]
- Center \(C_2 = (-3, -1)\)
- Radius \(r_2 = 3\)

### Step 2: Calculate the distance between the centers of the circles
Using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of the centers:
\[ d = \sqrt{((-3) - 3)^2 + ((-1) - 0)^2} = \sqrt{(-6)^2 + (-1)^2} = \sqrt{36 + 1} = \sqrt{37} \]

### Step 3: Compare the distance with the sum of the radii
- Sum of the radii:
\[ r_1 + r_2 = 3 + 3 = 6 \]
- Since \(d = \sqrt{37}\) and \(\sqrt{37} \approx 6.08\), we have:
\[ d > r_1 + r_2 \]

### Step 4: Determine the number of common tangents
Since the distance between the centers \(d\) is greater than the sum of the radii \(r_1 + r_2\), the circles do not intersect or touch. In this case, the number of common tangents is 4 (2 external and 2 internal).

### Conclusion
Thus, the number of common tangents that can be drawn to the circles is **4**.

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