The common tangent to the circles `x^(2)+y^(2) = 4 and x^(2) + y^(2) + 6x + 8y - 24 = 0` also passes through the point

Options

A. (6, -2)
B. (4, -2)
C. (-6, 4)
D. (-4, 6)

Explanation

To solve the problem of finding the common tangent to the circles given by the equations \(x^2 + y^2 = 4\) and \(x^2 + y^2 + 6x + 8y - 24 = 0\), and to determine which point it passes through, we can follow these steps:

### Step 1: Identify the centers and radii of the circles.

1. **First Circle**: The equation is \(x^2 + y^2 = 4\).
- Center \(C_1 = (0, 0)\)
- Radius \(r_1 = \sqrt{4} = 2\)

2. **Second Circle**: The equation is \(x^2 + y^2 + 6x + 8y - 24 = 0\).
- Rearranging this, we complete the square:
\[ x^2 + 6x + y^2 + 8y = 24 \]
\[ (x^2 + 6x + 9) + (y^2 + 8y + 16) = 49 \]
\[ (x + 3)^2 + (y + 4)^2 = 7^2 \]
- Center \(C_2 = (-3, -4)\)
- Radius \(r_2 = 7\)

### 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 centers:
\[ d = \sqrt{((-3) - 0)^2 + ((-4) - 0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 3: Determine the relationship between the circles.

Since \(d = 5\) (distance between centers) and \(r_1 + r_2 = 2 + 7 = 9\), we can see that:
- \(d < r_1 + r_2\) means the circles are internally tangent.

### Step 4: Find the equation of the common tangent.

The equation of the common tangent can be derived using the formula:
\[ S_1 - S_2 = 0 \]
Where \(S_1\) and \(S_2\) are the equations of the circles.

1. **For Circle 1**: \(S_1 = x^2 + y^2 - 4\)
2. **For Circle 2**: \(S_2 = x^2 + y^2 + 6x + 8y - 24\)

Setting up the equation:
\[ S_1 - S_2 = (x^2 + y^2 - 4) - (x^2 + y^2 + 6x + 8y - 24) = 0 \]
This simplifies to:
\[ -6x - 8y + 20 = 0 \]
Rearranging gives:
\[ 3x + 4y = 10 \]

### Step 5: Check which given point satisfies the tangent equation.

We need to check which point satisfies the equation \(3x + 4y = 10\). Let's check the points provided (assuming they are \( (6, -2), (2, 1), (0, 2), (-1, -1) \)).

1. **For point (6, -2)**:
\[ 3(6) + 4(-2) = 18 - 8 = 10 \quad \text{(satisfies)} \]

2. **For point (2, 1)**:
\[ 3(2) + 4(1) = 6 + 4 = 10 \quad \text{(satisfies)} \]

3. **For point (0, 2)**:
\[ 3(0) + 4(2) = 0 + 8 = 8 \quad \text{(does not satisfy)} \]

4. **For point (-1, -1)**:
\[ 3(-1) + 4(-1) = -3 - 4 = -7 \quad \text{(does not satisfy)} \]

### Conclusion

The points that satisfy the tangent equation are \( (6, -2) \) and \( (2, 1) \).