Find the equation of the common tangent of the following circles at their point of contact.
`x^2+y^2+10x-2y+22=0`
`x^2+y^2+2x-8y+8=0`

Answer

To find the equation of the common tangent of 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 given as:
1. \( S_1: x^2 + y^2 + 10x - 2y + 22 = 0 \)
2. \( S_2: x^2 + y^2 + 2x - 8y + 8 = 0 \)

### Step 2: Rearranging the equations

We can rearrange both equations to identify their centers and radii.

For \( S_1 \):
\[ x^2 + 10x + y^2 - 2y + 22 = 0 \]
Completing the square for \( x \) and \( y \):
\[ (x^2 + 10x + 25) + (y^2 - 2y + 1) = -22 + 25 + 1 \]
\[ (x + 5)^2 + (y - 1)^2 = 4 \]
This represents a circle with center \( (-5, 1) \) and radius \( 2 \).

For \( S_2 \):
\[ x^2 + 2x + y^2 - 8y + 8 = 0 \]
Completing the square for \( x \) and \( y \):
\[ (x^2 + 2x + 1) + (y^2 - 8y + 16) = -8 + 1 + 16 \]
\[ (x + 1)^2 + (y - 4)^2 = 9 \]
This represents a circle with center \( (-1, 4) \) and radius \( 3 \).

### Step 3: Finding the equation of the common tangent

To find the common tangent, we will use the relation \( S_1 - S_2 = 0 \).

Calculating \( S_1 - S_2 \):
\[ S_1 - S_2 = (x^2 + y^2 + 10x - 2y + 22) - (x^2 + y^2 + 2x - 8y + 8) = 0 \]
This simplifies to:
\[ 10x - 2y + 22 - 2x + 8y - 8 = 0 \]
Combining like terms:
\[ (10x - 2x) + (-2y + 8y) + (22 - 8) = 0 \]
\[ 8x + 6y + 14 = 0 \]
Dividing the entire equation by 2:
\[ 4x + 3y + 7 = 0 \]

### Final Answer

Thus, the equation of the common tangent is:
\[ \boxed{4x + 3y + 7 = 0} \]