How to find radius of a circle when equations of parallel tangents are given 
Equations of two parallel tangents to a circle are
  $2x-4y-9=0$ and $6x-12y+7=0$. How do I find the radius of the circle?

I have tried many methods, but still I  could not find the radius.
I know that the distance between two parallel tangents is equal to the diameter, but none of the contact point are given.
If I assume the centre to be $(a,b)$ and radius to be $r$ then when I apply the perpendicular distance of a line from a point I get two equations but I have three unknowns, so I cannot solve....any help will be appreciated 
 A: HINT


*

*Consider the line through the origin perpendicular to the tangents that is $y=-2x$

*Determine the intersections points P and Q between this perpendicular line and the tangents

*Then $2R=PQ$

A: Hint: The distance between two parallel lines, $y=mx + c_1$ and $y=mx+c_2$ is 
$$d = \frac{|c_1-c_2|}{\sqrt{1+m^2}}.$$
A: You need the distance between the two lines, which is the diameter. You should know how to compute the distance between a point and a line. Take an arbitrary (convenient) point on the most complicated looking line and find the distance from that point to the other line. Alternatively "draw a diagram" and "do the geometry (pythagoras)" - diagrams are surprisingly useful for this kind of problem..
A: Another way is:


*

*normalize the coefficients of $x,y$, i.e. find the unitary normal vector to the two lines
which thus shall be the same, and (if not) better to take with the same orientation
$$
{\bf n} = {1 \over {\sqrt {20} }}\left( {2, - 4} \right) = {1 \over {\sqrt {180} }}\left( {6, - 12} \right) = {1 \over {\sqrt 5 }}\left( {1, - 2} \right)
$$

*rewrite the equations accordingly
$$
{1 \over {\sqrt 5 }}x - {2 \over {\sqrt 5 }}y = {9 \over {2\sqrt 5 }}\quad \quad {1 \over {\sqrt 5 }}x - {2 \over {\sqrt 5 }}y =  - {7 \over {6\sqrt 5 }}
$$

*the above tells you that the two lines are at distance from the origin, measured in the orientation of $\bf n$, of
$ {9 \over {2\sqrt 5 }}$ and $ - {7 \over {6\sqrt 5 }}$, i.e. at ${17 \over {3\sqrt 5 }}$ from each other.
Then it is easy to write the equation of the middle line, and to put your circle with the center over that.
A: 
Given equations of the parallel lines
\begin{align}
y&=\tfrac12\,x-\tfrac9{4}
,\\
y&=\tfrac12\,x+\tfrac7{12}
,
\end{align}  
we can construct a triangle $ABC$
with two distinct points on one line 
and the third point on another line,
for example as
\begin{align} 
A&=(0,-\tfrac94),\quad
B=(\tfrac92,0),\quad
C=(0,\tfrac7{12})
.
\end{align}  
For this triangle we can find that
\begin{align} 
|AB|&=\sqrt{(B_x-A_x)^2+(B_y-A_y)^2}
=\tfrac{9\sqrt5}4
\end{align}
and the area can be found as
\begin{align} 
S_{\triangle ABC}
&=
\tfrac12((B_x-A_x)(C_y-A_y)
-(B_y-A_y)(C_x-A_x))
=\tfrac{51}8
.
\end{align}
Given that, we can find the height $|CD|$ of $\triangle ABD$
as
\begin{align}
|CD|&=\frac{2S_{\triangle ABC}}{|AB|}
=\frac{17\sqrt5}{15}
\end{align}
and the radius of the circle is then
\begin{align}
R&=\tfrac12|CD|=\frac{17\sqrt5}{30}
\approx 1.2671
.
\end{align}
