If the lines `3x-4y+4=0and6x-8y-7=0` are tangents to a circle ,then find the radius of the circle .

Options

A. `(3)/(4)`
B. `(4)/(3)`
C. `(1)/(4)`
D. `(7)/(4)`

Answer

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

### Step 1: Identify the equations of the lines
We have two lines:
1. \(3x - 4y + 4 = 0\) (Equation 1)
2. \(6x - 8y - 7 = 0\) (Equation 2)

### Step 2: Simplify the second line
We can simplify the second line by dividing the entire equation by 2:
\[ \frac{6x - 8y - 7}{2} = 0 \implies 3x - 4y - \frac{7}{2} = 0 \]
So, the second line can be rewritten as:
\[ 3x - 4y - \frac{7}{2} = 0 \quad (Equation 2') \]

### Step 3: Confirm that the lines are parallel
Both lines can be expressed in the form \(Ax + By + C = 0\):
- For Equation 1: \(A = 3\), \(B = -4\), \(C = 4\)
- For Equation 2': \(A = 3\), \(B = -4\), \(C = -\frac{7}{2}\)

Since the coefficients of \(x\) and \(y\) are the same in both equations, the lines are indeed parallel.

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

Calculating \(C_1 - C_2\):
\[ C_1 - C_2 = 4 - \left(-\frac{7}{2}\right) = 4 + \frac{7}{2} = \frac{8}{2} + \frac{7}{2} = \frac{15}{2} \]

Now, substituting into the distance formula:
\[ d = \frac{\left|\frac{15}{2}\right|}{\sqrt{3^2 + (-4)^2}} = \frac{\frac{15}{2}}{\sqrt{9 + 16}} = \frac{\frac{15}{2}}{\sqrt{25}} = \frac{\frac{15}{2}}{5} = \frac{15}{10} = \frac{3}{2} \]

### Step 5: Find the radius of the circle
Since the radius \(r\) of the circle is half the distance between the two parallel lines:
\[ r = \frac{d}{2} = \frac{\frac{3}{2}}{2} = \frac{3}{4} \]

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

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