The line `L_(1)-=3x-4y+1=0` touches the circles `C_(1) and C_(2)`. Centers of `C_(1)` and `C_(2)` are `A_(1)(1, 2)` and `A_(2)(3, 1)` respectively Then, the length (in units) of the transverse common tangent of `C_(1) and C_(2)` is equal to

Answer

To find the length of the transverse common tangent of the circles \( C_1 \) and \( C_2 \) with centers \( A_1(1, 2) \) and \( A_2(3, 1) \), we will follow these steps:

### Step 1: Identify the equation of the line and the centers of the circles.
The line is given by:
\[ L_1: 3x - 4y + 1 = 0 \]
The centers of the circles are:
\[ A_1(1, 2) \quad \text{and} \quad A_2(3, 1) \]

### Step 2: Calculate the distance from the centers of the circles to the line.
To find the radius of each circle, we will use the formula for the distance from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\):
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For the line \(3x - 4y + 1 = 0\), we have \(A = 3\), \(B = -4\), and \(C = 1\).

#### For Circle \(C_1\) (center \(A_1(1, 2)\)):
\[ d_1 = \frac{|3(1) - 4(2) + 1|}{\sqrt{3^2 + (-4)^2}} = \frac{|3 - 8 + 1|}{\sqrt{9 + 16}} = \frac{|-4|}{5} = \frac{4}{5} \]
Thus, the radius \(r_1\) of circle \(C_1\) is:
\[ r_1 = \frac{4}{5} \]

#### For Circle \(C_2\) (center \(A_2(3, 1)\)):
\[ d_2 = \frac{|3(3) - 4(1) + 1|}{\sqrt{3^2 + (-4)^2}} = \frac{|9 - 4 + 1|}{\sqrt{9 + 16}} = \frac{|6|}{5} = \frac{6}{5} \]
Thus, the radius \(r_2\) of circle \(C_2\) is:
\[ r_2 = \frac{6}{5} \]

### Step 3: Calculate the distance between the centers of the circles.
The distance \(d\) between the centers \(A_1(1, 2)\) and \(A_2(3, 1)\) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(3 - 1)^2 + (1 - 2)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]

### Step 4: Use the formula for the length of the transverse common tangent.
The length \(L\) of the transverse common tangent between two circles is given by:
\[ L = \sqrt{d^2 - (r_1 + r_2)^2} \]
Substituting the values we found:
\[ L = \sqrt{(\sqrt{5})^2 - \left(\frac{4}{5} + \frac{6}{5}\right)^2} = \sqrt{5 - \left(\frac{10}{5}\right)^2} = \sqrt{5 - 4} = \sqrt{1} = 1 \]

### Final Answer:
The length of the transverse common tangent of circles \(C_1\) and \(C_2\) is:
\[ \boxed{1} \]