The circles `x^(2)+y^(2)- 10x+ 4y - 20=0 and x^(2)+y^(2)+ 14x-6y+22=0` touch each other.

Answer

To determine if the two circles touch each other, 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. \( x^2 + y^2 - 10x + 4y - 20 = 0 \)
2. \( x^2 + y^2 + 14x - 6y + 22 = 0 \)

### Step 2: Identify the coefficients for each circle
For the first circle, we can identify:
- \( G_1 = -10 \)
- \( F_1 = 4 \)
- \( C_1 = -20 \)

For the second circle:
- \( G_2 = 14 \)
- \( F_2 = -6 \)
- \( C_2 = 22 \)

### Step 3: Find the center and radius of each circle
The center \((h, k)\) and radius \(r\) of a circle given by the general equation \(x^2 + y^2 + Gx + Fy + C = 0\) can be calculated as follows:
- Center: \( C = \left(-\frac{G}{2}, -\frac{F}{2}\right) \)
- Radius: \( r = \sqrt{\left(-\frac{G}{2}\right)^2 + \left(-\frac{F}{2}\right)^2 - C} \)

**For Circle 1:**
- Center \( C_1 = \left(-\frac{-10}{2}, -\frac{4}{2}\right) = (5, -2) \)
- Radius \( R_1 = \sqrt{\left(-\frac{-10}{2}\right)^2 + \left(-\frac{4}{2}\right)^2 - (-20)} \)
\[ R_1 = \sqrt{(5)^2 + (-2)^2 + 20} = \sqrt{25 + 4 + 20} = \sqrt{49} = 7 \]

**For Circle 2:**
- Center \( C_2 = \left(-\frac{14}{2}, -\frac{-6}{2}\right) = (-7, 3) \)
- Radius \( R_2 = \sqrt{\left(-\frac{14}{2}\right)^2 + \left(-\frac{-6}{2}\right)^2 - 22} \)
\[ R_2 = \sqrt{(-7)^2 + (3)^2 - 22} = \sqrt{49 + 9 - 22} = \sqrt{36} = 6 \]

### Step 4: Calculate the distance between the centers of the circles
The distance \(d\) between the centers \(C_1\) and \(C_2\) can be calculated using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the values:
\[ d = \sqrt{(-7 - 5)^2 + (3 - (-2))^2} = \sqrt{(-12)^2 + (5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]

### Step 5: Check the condition for the circles to touch each other
The circles touch each other if the distance between their centers \(d\) is equal to the sum of their radii \(R_1 + R_2\) or the absolute difference of their radii \(|R_1 - R_2|\).

Here:
- \( R_1 + R_2 = 7 + 6 = 13 \)
- \( |R_1 - R_2| = |7 - 6| = 1 \)

Since \(d = 13\) is equal to \(R_1 + R_2\), the circles touch each other externally.

### Final Conclusion
The circles touch each other externally.

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