The number of common tangents to the circles `x^(2)+y^(2)-4x-6y-12=0` and `x^(2)+y^(2)+6x+18y+26=0`, is

Options

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

Explanation

To find the number of common tangents to the given circles, we need to analyze their equations and properties step by step.

### Step 1: Identify the equations of the circles
The equations of the circles are:
1. \( x^2 + y^2 - 4x - 6y - 12 = 0 \)
2. \( x^2 + y^2 + 6x + 18y + 26 = 0 \)

### Step 2: Rewrite the equations in standard form
We will convert each equation into the standard form of a circle, which is \( (x - h)^2 + (y - k)^2 = r^2 \).

**For the first circle:**
1. Rearranging the first equation:
\[ x^2 - 4x + y^2 - 6y = 12 \]
2. 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 \( r_1 = 5 \).

**For the second circle:**
1. Rearranging the second equation:
\[ x^2 + 6x + y^2 + 18y = -26 \]
2. Completing the square:
\[ (x^2 + 6x + 9) + (y^2 + 18y + 81) = -26 + 9 + 81 \]
\[ (x + 3)^2 + (y + 9)^2 = 64 \]
Thus, the center \( C_2 \) is \( (-3, -9) \) and the radius \( r_2 = 8 \).

### Step 3: Calculate the distance between the centers
Now we calculate the distance \( d \) between the centers \( C_1(2, 3) \) and \( C_2(-3, -9) \):
\[ d = \sqrt{(2 - (-3))^2 + (3 - (-9))^2} \]
\[ d = \sqrt{(2 + 3)^2 + (3 + 9)^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \]

### Step 4: Compare the distance with the sum of the radii
Now, we find the sum of the radii:
\[ r_1 + r_2 = 5 + 8 = 13 \]

### Step 5: Determine the relationship between the circles
Since the distance between the centers \( d = 13 \) is equal to the sum of the radii \( r_1 + r_2 = 13 \), the circles are externally tangent to each other.

### Step 6: Conclusion on the number of common tangents
When two circles are externally tangent, they have:
- 3 common tangents: 2 external tangents and 1 internal tangent.

Thus, the number of common tangents to the given circles is **3**.

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