If `2x-4y=9` and `6x-12y+7=0` are parallel tangents to 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 solve the problem, we need to find the radius of the circle given that the lines \(2x - 4y = 9\) and \(6x - 12y + 7 = 0\) are parallel tangents to the circle.

### Step-by-Step Solution:

1. **Convert the equations of the lines into standard form**:
- The first line is given as \(2x - 4y = 9\). Dividing the entire equation by 2 gives:
\[ x - 2y = \frac{9}{2} \quad \text{(Equation 1)} \]
- The second line is given as \(6x - 12y + 7 = 0\). Dividing the entire equation by 6 gives:
\[ x - 2y = -\frac{7}{6} \quad \text{(Equation 2)} \]

2. **Identify coefficients**:
- From both equations, we can identify \(a = 1\), \(b = -2\), \(c_1 = -\frac{9}{2}\), and \(c_2 = \frac{7}{6}\).

3. **Use the formula for the distance between 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:
\[ d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} \]
- Applying this formula to our lines:
\[ d = \frac{\left|-\frac{9}{2} - \left(-\frac{7}{6}\right)\right|}{\sqrt{1^2 + (-2)^2}} \]

4. **Calculate the numerator**:
- First, find a common denominator for \(-\frac{9}{2}\) and \(-\frac{7}{6}\):
\[ -\frac{9}{2} = -\frac{27}{6} \]
- Thus, we have:
\[ d = \frac{\left|-\frac{27}{6} + \frac{7}{6}\right|}{\sqrt{1 + 4}} = \frac{\left|-\frac{27 - 7}{6}\right|}{\sqrt{5}} = \frac{\left|-\frac{20}{6}\right|}{\sqrt{5}} = \frac{\frac{20}{6}}{\sqrt{5}} = \frac{10}{3\sqrt{5}} \]

5. **Relate the distance to the radius**:
- Since the distance \(d\) between the two tangents is equal to \(2r\) (where \(r\) is the radius of the circle), we have:
\[ 2r = \frac{10}{3\sqrt{5}} \]

6. **Solve for the radius \(r\)**:
- Dividing both sides by 2 gives:
\[ r = \frac{10}{6\sqrt{5}} = \frac{5}{3\sqrt{5}} \]

7. **Rationalize the denominator**:
- To rationalize the denominator:
\[ r = \frac{5\sqrt{5}}{15} = \frac{\sqrt{5}}{3} \]

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