If `L _(1) : 4x - 3y + 7=0 and L _(2) : 8x - 6y -1 =0` are two tangents to a circle C, length of whose diameters is d, then d is equal to

Answer

To find the diameter \( d \) of the circle given the tangents \( L_1: 4x - 3y + 7 = 0 \) and \( L_2: 8x - 6y - 1 = 0 \), we will follow these steps:

### Step 1: Identify the equations of the lines
The equations of the tangents are:
- \( L_1: 4x - 3y + 7 = 0 \)
- \( L_2: 8x - 6y - 1 = 0 \)

### Step 2: Simplify the second line
We can simplify \( L_2 \) by dividing the entire equation by 2:
\[ L_2: \frac{8x}{2} - \frac{6y}{2} - \frac{1}{2} = 0 \implies 4x - 3y - \frac{1}{2} = 0 \]

### Step 3: Compare the slopes of the lines
The slope of a line given by the equation \( ax + by + c = 0 \) is given by \( -\frac{a}{b} \).
- For \( L_1 \):
\[ \text{slope } M_1 = -\frac{4}{-3} = \frac{4}{3} \]
- For \( L_2 \):
\[ \text{slope } M_2 = -\frac{4}{-3} = \frac{4}{3} \]

Since both slopes are equal, the lines are parallel.

### Step 4: Use the formula for the distance between two parallel lines
The distance \( d \) between two parallel lines \( Ax + By + C_1 = 0 \) and \( Ax + By + C_2 = 0 \) is given by:
\[ d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} \]
Here, for our lines:
- From \( L_1 \): \( C_1 = 7 \)
- From \( L_2 \): \( C_2 = -\frac{1}{2} \)

### Step 5: Calculate the distance
Substituting into the distance formula:
\[ d = \frac{|-\frac{1}{2} - 7|}{\sqrt{4^2 + (-3)^2}} = \frac{|-\frac{1}{2} - \frac{14}{2}|}{\sqrt{16 + 9}} = \frac{|\frac{-15}{2}|}{\sqrt{25}} = \frac{\frac{15}{2}}{5} = \frac{15}{10} = \frac{3}{2} \]

### Step 6: Conclusion
Since the distance between the two tangents is equal to the diameter of the circle, we have:
\[ d = \frac{3}{2} \]

Thus, the diameter \( d \) of the circle is \( \frac{3}{2} \).

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