The circle `x^(2)+y^(2)-4x-6y-12=0, x^(2)+y^(2)+6x-8y+21=0` are

Options

A. intersection
B. toching exterally
C. touching internally
D. one is lying inside the other

Answer

To solve the problem of determining the relationship between the two circles given by the equations \( x^2 + y^2 - 4x - 6y - 12 = 0 \) and \( x^2 + y^2 + 6x - 8y + 21 = 0 \), we will follow these steps:

### Step 1: Rewrite the Circle Equations
We need to rewrite both equations in the standard form of a circle, which is \((x - h)^2 + (y - k)^2 = r^2\).

1. **For Circle 1:**
\[ x^2 + y^2 - 4x - 6y - 12 = 0 \]
Rearranging gives:
\[ x^2 - 4x + y^2 - 6y = 12 \]
Completing the square:
\[ (x^2 - 4x + 4) + (y^2 - 6y + 9) = 12 + 4 + 9 \]
\[ (x - 2)^2 + (y - 3)^2 = 25 \]
Thus, the center \(C_1\) is \((2, 3)\) and the radius \(r_1 = 5\).

2. **For Circle 2:**
\[ x^2 + y^2 + 6x - 8y + 21 = 0 \]
Rearranging gives:
\[ x^2 + 6x + y^2 - 8y = -21 \]
Completing the square:
\[ (x^2 + 6x + 9) + (y^2 - 8y + 16) = -21 + 9 + 16 \]
\[ (x + 3)^2 + (y - 4)^2 = 4 \]
Thus, the center \(C_2\) is \((-3, 4)\) and the radius \(r_2 = 2\).

### Step 2: Calculate the Distance Between the Centers
Now, we will calculate the distance \(d\) between the centers \(C_1\) and \(C_2\).

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{((-3) - 2)^2 + (4 - 3)^2} \]
\[ d = \sqrt{(-5)^2 + (1)^2} = \sqrt{25 + 1} = \sqrt{26} \]

### Step 3: Compare Distances and Radii
To determine the relationship between the circles, we compare the distance \(d\) with the sum and difference of the radii:

1. **Sum of the Radii:**
\[ r_1 + r_2 = 5 + 2 = 7 \]

2. **Difference of the Radii:**
\[ |r_1 - r_2| = |5 - 2| = 3 \]

### Step 4: Determine the Relationship
Now we analyze the distances:
- \(d = \sqrt{26} \approx 5.1\)
- \(r_1 + r_2 = 7\)
- \(r_1 - r_2 = 3\)

Since \(3 < \sqrt{26} < 7\), it indicates that the circles are touching internally.

### Conclusion
The two circles touch internally.

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