The circles `x^(2)+y^(2)-12x+8y+48=0, x^(2)+y^(2)-4x+2y-4=0` are

Options

A. intersecting
B. touching externally
C. touching internally
D. one is lying inside the other

Answer

To determine the nature of the two circles given by the equations \(x^2 + y^2 - 12x + 8y + 48 = 0\) and \(x^2 + y^2 - 4x + 2y - 4 = 0\), we will follow these steps:

### Step 1: Rewrite the equations in standard form
We start by rewriting the equations of the circles in standard form \((x - h)^2 + (y - k)^2 = r^2\).

**Circle 1:**
\[ x^2 + y^2 - 12x + 8y + 48 = 0 \]
Rearranging gives:
\[ x^2 - 12x + y^2 + 8y + 48 = 0 \]
Completing the square for \(x\) and \(y\):
\[ (x^2 - 12x + 36) + (y^2 + 8y + 16) = -48 + 36 + 16 \]
This simplifies to:
\[ (x - 6)^2 + (y + 4)^2 = 4 \]
Thus, the center \(C_1\) of Circle 1 is \((6, -4)\) and the radius \(r_1 = 2\).

**Circle 2:**
\[ x^2 + y^2 - 4x + 2y - 4 = 0 \]
Rearranging gives:
\[ x^2 - 4x + y^2 + 2y - 4 = 0 \]
Completing the square for \(x\) and \(y\):
\[ (x^2 - 4x + 4) + (y^2 + 2y + 1) = 4 + 4 + 1 \]
This simplifies to:
\[ (x - 2)^2 + (y - 1)^2 = 9 \]
Thus, the center \(C_2\) of Circle 2 is \((2, 1)\) and the radius \(r_2 = 3\).

### Step 2: Calculate the distance between the centers
Now, we calculate the distance \(d\) between the centers \(C_1\) and \(C_2\):
\[ d = \sqrt{(6 - 2)^2 + (-4 - 1)^2} = \sqrt{4^2 + (-5)^2} = \sqrt{16 + 25} = \sqrt{41} \]

### Step 3: Compare the distance with the sum of the radii
Next, we find the sum of the radii:
\[ r_1 + r_2 = 2 + 3 = 5 \]

### Step 4: Determine the nature of the circles
Now we compare the distance \(d\) with the sum of the radii:
- If \(d < r_1 + r_2\), the circles intersect at two points.
- If \(d = r_1 + r_2\), the circles touch externally.
- If \(d > r_1 + r_2\), the circles are separate.

In this case:
\[ \sqrt{41} \approx 6.4 > 5 \]
Thus, the circles are separate.

### Final Conclusion
The two circles are separate.