To solve the problem, we need to find the absolute value of \((\alpha + \beta)(\gamma + \delta)\) where \((\alpha, \beta)\) and \((\gamma, \delta)\) are the centers of two circles that touch each other at the point (1, 2) and have equal radii of 5 cm. Additionally, we have a common tangent given by the equation \(4x + 3y = 10\).
### Step-by-step Solution:
1. **Identify the centers of the circles**:
The circles touch each other at the point (1, 2). Since both circles have equal radii of 5 cm, the distance between their centers will be \(2 \times 5 = 10\) cm.
2. **Find the slope of the tangent**:
The equation of the tangent is given as \(4x + 3y = 10\). We can rewrite this in slope-intercept form:
\[
3y = -4x + 10 \implies y = -\frac{4}{3}x + \frac{10}{3}
\]
The slope of the tangent line is \(-\frac{4}{3}\).
3. **Determine the slope of the line connecting the centers**:
Since the line connecting the centers of the circles (let's call it \(C_1C_2\)) is perpendicular to the tangent, the slope of \(C_1C_2\) will be the negative reciprocal of the slope of the tangent:
\[
m_{C_1C_2} = \frac{3}{4}
\]
4. **Use the point-slope form to find the centers**:
Let the center of the first circle be \((\alpha, \beta)\) and the center of the second circle be \((\gamma, \delta)\). The line connecting these centers can be expressed in point-slope form as:
\[
y - 2 = \frac{3}{4}(x - 1)
\]
Rearranging gives us:
\[
3x - 4y + 5 = 0
\]
5. **Finding the coordinates of the centers**:
The distance between the centers is 10 cm. We can express the centers in terms of a parameter \(\theta\):
\[
C_1: (1 - 5 \cos \theta, 2 - 5 \sin \theta)
\]
\[
C_2: (1 + 5 \cos \theta, 2 + 5 \sin \theta)
\]
6. **Calculate \(\alpha\), \(\beta\), \(\gamma\), and \(\delta\)**:
Using the trigonometric identities:
\[
\cos \theta = \frac{4}{5}, \quad \sin \theta = \frac{3}{5}
\]
We substitute these values:
\[
C_1: (1 - 5 \cdot \frac{4}{5}, 2 - 5 \cdot \frac{3}{5}) = (1 - 4, 2 - 3) = (-3, -1)
\]
\[
C_2: (1 + 5 \cdot \frac{4}{5}, 2 + 5 \cdot \frac{3}{5}) = (1 + 4, 2 + 3) = (5, 5)
\]
Thus, we have:
\[
\alpha = -3, \quad \beta = -1, \quad \gamma = 5, \quad \delta = 5
\]
7. **Calculate \((\alpha + \beta)(\gamma + \delta)\)**:
\[
\alpha + \beta = -3 - 1 = -4
\]
\[
\gamma + \delta = 5 + 5 = 10
\]
Therefore:
\[
(\alpha + \beta)(\gamma + \delta) = (-4)(10) = -40
\]
8. **Find the absolute value**:
\[
\text{abs}((\alpha + \beta)(\gamma + \delta)) = | -40 | = 40
\]
### Final Answer:
\[
\text{abs}((\alpha + \beta)(\gamma + \delta)) = 40
\]
