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 circles given by the equations \(x^2 + y^2 - 4x - 6y - 12 = 0\) and \(x^2 + y^2 + 6x + 18y + 26 = 0\), 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:**
\[ x^2 + y^2 - 4x - 6y - 12 = 0 \]
Rearranging gives:
\[ (x^2 - 4x) + (y^2 - 6y) = 12 \]
Completing the square:
\[ (x - 2)^2 - 4 + (y - 3)^2 - 9 = 12 \]
\[ (x - 2)^2 + (y - 3)^2 = 25 \]
Thus, the center \(C_1\) is \((2, 3)\) and the radius \(r_1 = 5\).

**Circle 2:**
\[ x^2 + y^2 + 6x + 18y + 26 = 0 \]
Rearranging gives:
\[ (x^2 + 6x) + (y^2 + 18y) = -26 \]
Completing the square:
\[ (x + 3)^2 - 9 + (y + 9)^2 - 81 = -26 \]
\[ (x + 3)^2 + (y + 9)^2 = 64 \]
Thus, the center \(C_2\) is \((-3, -9)\) and the radius \(r_2 = 8\).

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

### Step 3: Determine the relationship between the circles
We compare the distance \(d\) with the sum and difference of the radii:
- \(r_1 + r_2 = 5 + 8 = 13\)
- \(r_1 - r_2 = 5 - 8 = -3\) (which is negative, so we ignore this)

Since \(d = r_1 + r_2\), the circles touch each other externally.

### Step 4: Conclusion on the number of common tangents
When two circles touch externally, they have:
- 2 external tangents
- 1 internal tangent (the radical axis)

Thus, the total number of common tangents is:
\[ \text{Total common tangents} = 2 + 1 = 3 \]

### Final Answer
The number of common tangents to the circles is **3**.

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