Let $L$ be a line and $P$ be a point outside of the line. Given a radius $r$, construct a circle with radius $r$ that passes through $P$ and is tangent to the given line.
I know the center of the circle asked must be a point in the circle with radius $r$ and center $P$, but I don't know where to take it from there.
On
General equation of the circle is
$$ (x-a)^2 +(y-b)^2 =r^2 $$
where $(a,b)$ is the centre of the circle. The radius is given. We need to find the centre $(a,b)$.
We are given a point $P=(p,q)$ that belongs to the circle, thus
$$ (p-a)^2+(q-b)^2=r^2 $$
This gives one equation for $(a,b)$. The second equation is given by the tangent line. Let $Ax+By+C=0$ be the tangent line. The condition for it to be tangent means that it has two coinciding common points with the circle. Or, alternatively, the equations:
$$(x-a)^2+(y-b)^2=r^2$$
$$Ax+By+C=0$$
have two coinciding solutions. This information should be enough to find $(a,b)$ and solve the problem.
NB: If you don't mind using a bit of calculus, it can help with the solution. Gradient to the circle at point $(x,y)$ is a vector $\{x-a,y-b\}$. This vector is perpendicular to the tangent line and, thus, is parallel to vector $\{A,B\}$. This gives a necessary, but not sufficient condition for the line to be tangent.
This is actually easier than the analytical solution. I will describe the construction process, I hope it will be clear enough to be reproducible.
We are given point $P$ and line $l$ on the plane and the length of the radius $r$. The centre $O$ of the circle must lie: (a) in the same semi-plane as $P$ relative to $l$, (b) on the line $l_r$ parallel to $l$ and at the distance $r$ from $l$ and (c) at distance $r$ from $P$.
Line $l'$ is constructed by constructing two perpendicular lines to $l$ at some points $A$ and $B$ and constructing segments $AA'=BB'=r$ to the semi-plane containing $P$, then $A'B'=l'$.
To satisfy (c) the centre must lie somewhere on the circle of radius $r$ with centre at $P$. Find the intersection of the circle with $l'$ and you get the centre $O$.
NB: depending on the choice of $P$, $l$ and $r$ there can be either two solutions, one or none. Can you find the condition when there are the solutions?
A circle is the set of points with distance $r$ from the center. You want to find the center that is distant of $r$ from $P$ and from $L$.
To be tangent to $L$, your circle must have its center at a distance of $r$ from $L$. This constraints the center to be onto one the two lines parallel to $L$ at a distance of $r$ from $L$.
To contains $P$, your circle must have its center at a distance of $r$ from $P$. This constraints the center to be onto the circle or radius $r$ and center $P$.
Hence your center must be at the intersection of the two constraints.