TITLE: P and Q are 4 units apart and on the same side of variable straight line which touches a circle
QUESTION [1 upvotes]: My approach: I took a variable line ax+by+c=0 and took random points like (2,0) and (6,0). The I applied the formula for perpendicular distance and put it in the condition that PM+3QN=4.
but this way there are three variable and one equation by which I am unable to get the equation of the line and satisfy it with the tangency condition of the circle that perpendicular distance from centre to this line is equal to radius but this way more and more number of variables are introduced and lesser number of equations.

REPLY [2 votes]: If you call $\theta$ the angle $\angle MPQ$, then you have
$$\overline{PM} = 4-3\cdot\overline{QN} = 4\cos\theta-\overline{QN},$$
which gives you 
\begin{equation}\overline{QN} = 1-\cos\theta\tag{1}\label{uno}\end{equation}
and consequently
\begin{equation}\overline{PM} = 1+3\cos\theta.\tag{2}\label{due}\end{equation}
Then it is useful to put $P$ in the origin of the axes.

Note that
$$\overline{PH} = 3\cos\theta$$
and, thus, from \eqref{due},
$$\overline{HM} = 1.$$
Then if you draw the line $HK$ perpendicular to $PM$ you get
$$\overline{KN} = 1,$$
and 
$$\overline{KQ} = \cos\theta.$$
Again, as in \eqref{uno},
$$\overline{QN} = 1-\cos\theta.$$
As a consequence, the distance of the line $MN$ from $G(3,0)$ will be constant, and the line will be tangent to the circle centered in $G$ with radius $1$.
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