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 between the two circles \( C_1 \) and \( C_2 \) with centers \( A_1(1, 2) \) and \( A_2(3, 1) \), and given that the line \( L_1: 3x - 4y + 1 = 0 \) touches both circles, we can follow these steps:

### Step 1: Find the distance from the center of each circle to the line
The formula for the distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:

\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

For the line \( 3x - 4y + 1 = 0 \), we have \( A = 3 \), \( B = -4 \), and \( C = 1 \).

#### Finding \( r_1 \) (radius of circle \( C_1 \)):
For center \( A_1(1, 2) \):

\[ r_1 = \frac{|3(1) - 4(2) + 1|}{\sqrt{3^2 + (-4)^2}} = \frac{|3 - 8 + 1|}{\sqrt{9 + 16}} = \frac{|-4|}{5} = \frac{4}{5} \]

#### Finding \( r_2 \) (radius of circle \( C_2 \)):
For center \( A_2(3, 1) \):

\[ r_2 = \frac{|3(3) - 4(1) + 1|}{\sqrt{3^2 + (-4)^2}} = \frac{|9 - 4 + 1|}{5} = \frac{|6|}{5} = \frac{6}{5} \]

### Step 2: Calculate the distance between the centers \( A_1 \) and \( A_2 \)
Using the distance formula:

\[ d(A_1, A_2) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(3 - 1)^2 + (1 - 2)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]

### Step 3: Use the formula for the length of the transverse common tangent
The length \( L \) of the transverse common tangent is given by:

\[ L = \sqrt{d(A_1, A_2)^2 - (r_1 + r_2)^2} \]

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 \( 1 \) unit.

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