If `2x-4y=9and6x-12y+7=0` are common tangents to a circle , then radius of the circle is

Options

A. `(sqrt(3))/(5)`
B. `(17)/(6sqrt(5))`
C. `(sqrt(2))/(3)`
D. `(17)/(3sqrt(5))`

Answer

To find the radius of the circle given that the lines \(2x - 4y = 9\) and \(6x - 12y + 7 = 0\) are common tangents, we can follow these steps:

### Step 1: Convert the equations of the lines into slope-intercept form
1. **For the first line \(2x - 4y = 9\):**
\[ -4y = -2x + 9 \implies y = \frac{1}{2}x - \frac{9}{4} \]
Here, the slope \(m_1 = \frac{1}{2}\) and \(c_1 = -\frac{9}{4}\).

2. **For the second line \(6x - 12y + 7 = 0\):**
\[ -12y = -6x - 7 \implies y = \frac{1}{2}x + \frac{7}{12} \]
Here, the slope \(m_2 = \frac{1}{2}\) and \(c_2 = \frac{7}{12}\).

### Step 2: Verify that the lines are parallel
Since both lines have the same slope \(m_1 = m_2 = \frac{1}{2}\), they are indeed parallel.

### Step 3: 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:
\[ d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} \]
For our lines:
- The first line can be rewritten as \(2x - 4y - 9 = 0\) (where \(C_1 = -9\)).
- The second line can be rewritten as \(6x - 12y + 7 = 0\) (where \(C_2 = 7\)).

To use the formula, we need to express both lines in the same form. We can multiply the first line by 3 to match the coefficients:
\[ 6x - 12y - 27 = 0 \quad (C_1 = -27) \]

Now we have:
- \(C_1 = -27\)
- \(C_2 = 7\)

### Step 4: Substitute into the distance formula
Now we can calculate the distance:
\[ d = \frac{|7 - (-27)|}{\sqrt{6^2 + (-12)^2}} = \frac{|7 + 27|}{\sqrt{36 + 144}} = \frac{34}{\sqrt{180}} = \frac{34}{6\sqrt{5}} = \frac{17}{3\sqrt{5}} \]

### Step 5: Relate the distance to the radius
Since the distance \(d\) between the two tangents is equal to the diameter of the circle, the radius \(r\) is:
\[ r = \frac{d}{2} = \frac{17}{6\sqrt{5}} \]

### Final Answer
The radius of the circle is:
\[ \boxed{\frac{17}{6\sqrt{5}}} \]