Find the number of common tangents to the circles `x^(2)+y^(2)-8x+2y+8=0andx^(2)+y^(2)-2x-6y-15=0`.

Answer

To find the number of common tangents to the given circles, we will follow these steps:

### Step 1: Write the equations of the circles in standard form.
The equations of the circles are:
1. \( x^2 + y^2 - 8x + 2y + 8 = 0 \)
2. \( x^2 + y^2 - 2x - 6y - 15 = 0 \)

### Step 2: Identify the center and radius of the first circle.
For the first circle:
- The equation can be rewritten as:
\[ (x^2 - 8x) + (y^2 + 2y) + 8 = 0 \]
Completing the square:
\[ (x - 4)^2 - 16 + (y + 1)^2 - 1 + 8 = 0 \]
\[ (x - 4)^2 + (y + 1)^2 - 9 = 0 \]
Thus, the center \( C_1 \) is \( (4, -1) \) and the radius \( R_1 \) is \( 3 \) (since \( R_1 = \sqrt{9} \)).

### Step 3: Identify the center and radius of the second circle.
For the second circle:
- The equation can be rewritten as:
\[ (x^2 - 2x) + (y^2 - 6y) - 15 = 0 \]
Completing the square:
\[ (x - 1)^2 - 1 + (y - 3)^2 - 9 - 15 = 0 \]
\[ (x - 1)^2 + (y - 3)^2 - 25 = 0 \]
Thus, the center \( C_2 \) is \( (1, 3) \) and the radius \( R_2 \) is \( 5 \) (since \( R_2 = \sqrt{25} \)).

### Step 4: Calculate the distance between the centers of the circles.
The distance \( d \) between the centers \( C_1 \) and \( C_2 \) is given by the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{(1 - 4)^2 + (3 - (-1))^2} = \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 5: Determine the number of common tangents.
Now we compare \( d \) with \( R_1 \) and \( R_2 \):
- \( R_1 + R_2 = 3 + 5 = 8 \)
- \( d = 5 \)

Since \( R_1 + R_2 > d \) and \( |R_1 - R_2| < d \) (i.e., \( 2 < 5 \)), the circles intersect at two distinct points.

### Conclusion:
The number of common tangents to the circles is **2**.

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