If the lines 3 x - 4y + 4 = 0 and 6x - 8y - 7 = 0 are tangents to a circle , then the radius of the circle is

Options

A. `(3)/(4)` units
B. `(3)/(2)` units
C. 3 units
D. `(2)/(3)` units

Answer

To find the radius of the circle given that the lines \(3x - 4y + 4 = 0\) and \(6x - 8y - 7 = 0\) are tangents to the circle, we can follow these steps:

### Step 1: Write the equations of the tangents
The equations of the tangents are:
- \(T_1: 3x - 4y + 4 = 0\)
- \(T_2: 6x - 8y - 7 = 0\)

### Step 2: Simplify the second tangent
We can simplify \(T_2\) by dividing the entire equation by 2:
\[ T_2: 3x - 4y - \frac{7}{2} = 0 \]

### Step 3: Identify coefficients
From the equations of the tangents, we can identify:
- For \(T_1\): \(A = 3\), \(B = -4\), \(C_1 = 4\)
- For \(T_2\): \(C_2 = -\frac{7}{2}\)

### Step 4: Check if the lines are parallel
The slopes of both lines can be determined from their equations. Since both lines have the same coefficients for \(x\) and \(y\) (i.e., \(3\) and \(-4\)), they are parallel.

### Step 5: Use the formula for the distance between two parallel lines
The distance \(D\) between two parallel lines given by \(Ax + By + C_1 = 0\) and \(Ax + By + C_2 = 0\) is given by:
\[ D = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} \]

Substituting the values we have:
\[ D = \frac{\left| -\frac{7}{2} - 4 \right|}{\sqrt{3^2 + (-4)^2}} = \frac{\left| -\frac{7}{2} - \frac{8}{2} \right|}{\sqrt{9 + 16}} = \frac{\left| -\frac{15}{2} \right|}{\sqrt{25}} = \frac{\frac{15}{2}}{5} \]

### Step 6: Simplify the distance
\[ D = \frac{15}{2} \cdot \frac{1}{5} = \frac{15}{10} = \frac{3}{2} \]

### Step 7: Relate the distance to the diameter of the circle
The distance between the two tangents is equal to the diameter of the circle:
\[ \text{Diameter} = D = \frac{3}{2} \]

### Step 8: Find the radius of the circle
The radius \(r\) of the circle is half of the diameter:
\[ r = \frac{D}{2} = \frac{3/2}{2} = \frac{3}{4} \]

### Final Answer
The radius of the circle is \(\frac{3}{4}\) units.

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