How to find radius of a circle when equations of parallel tangents are given

Question

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

Answer 1

Hint: The distance between two parallel lines, $y=mx + c_1$ and $y=mx+c_2$ is $$d = \frac{|c_1-c_2|}{\sqrt{1+m^2}}.$$

Answer 2

HINT

1. Consider the line through the origin perpendicular to the tangents that is $y=-2x$
2. Determine the intersections points P and Q between this perpendicular line and the tangents
3. Then $2R=PQ$

Answer 3

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..

Answer 4

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.

Answer 5

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}