# 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

• Neither the points of contact nor the centre of the circle is needed. Hint: Start by choosing a point on one of the lines. – Henrik May 13 '18 at 21:10

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

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 $2R=PQ$

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..

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.

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

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}