The point lying on common tangent to the circles `x^(2)+y^(2)=4` and `x^(2)+y^(2)+6x+8y-24=0` is (1) (4,-2) (2) (-6,4) (3) (6,-2) (4) (-4,6)

Options

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

Answer

To find the point lying on the common tangent to the circles given by the equations \(x^2 + y^2 = 4\) and \(x^2 + y^2 + 6x + 8y - 24 = 0\), we will follow these steps:

### Step 1: Identify the equations of the circles
The first circle is given by:
\[ x^2 + y^2 = 4 \]
This represents a circle centered at \((0, 0)\) with a radius of \(2\).

The second circle can be rewritten by completing the square:
\[ x^2 + y^2 + 6x + 8y - 24 = 0 \]
Rearranging gives:
\[ x^2 + 6x + y^2 + 8y = 24 \]
Completing the square for \(x\):
\[ (x + 3)^2 - 9 \]
Completing the square for \(y\):
\[ (y + 4)^2 - 16 \]
Thus, the equation becomes:
\[ (x + 3)^2 + (y + 4)^2 - 25 = 0 \]
This represents a circle centered at \((-3, -4)\) with a radius of \(5\).

### Step 2: Find the equation of the common tangent
The formula for the equation of the common tangent to two circles is given by:
\[ s_1 - s_2 = 0 \]
where \(s_1\) and \(s_2\) are the equations of the circles.

For the first circle:
\[ s_1 = x^2 + y^2 - 4 \]

For the second circle:
\[ s_2 = x^2 + y^2 + 6x + 8y - 24 \]

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:
\[ 6x + 8y = 20 \]
Dividing by 2:
\[ 3x + 4y = 10 \]

### Step 3: Check the given points
We need to check which of the given points satisfies the equation \(3x + 4y = 10\).

1. **Point (4, -2)**:
\[ 3(4) + 4(-2) = 12 - 8 = 4 \quad \text{(not equal to 10)} \]

2. **Point (-6, 4)**:
\[ 3(-6) + 4(4) = -18 + 16 = -2 \quad \text{(not equal to 10)} \]

3. **Point (6, -2)**:
\[ 3(6) + 4(-2) = 18 - 8 = 10 \quad \text{(equal to 10)} \]

4. **Point (-4, 6)**:
\[ 3(-4) + 4(6) = -12 + 24 = 12 \quad \text{(not equal to 10)} \]

### Conclusion
The point that lies on the common tangent to the circles is:
\[ \boxed{(6, -2)} \]