# PLC: Apollonius Problem with a point, a line, and a circle

Below I give an Euclidean construction to a variant of the Problem of Apollonius:

Find a circle tangent to a given line $m$, circle $(A)$, and passing through a given point $B$:

(The applet below illustrates the construction. Points $A, B$ and line $m$ are draggable. This permits changing the configuration, but within certain limitations.)

Solution

### Construction

Find a circle tangent to a given line $m$, circle $(A)$, and passing through a given point $B$:

Let's analyze the problem assuming that a solution has been found. Assume there is a circle through $O$ through $B$ that touches $m$ at $P$ and $(A)$ at $L:$

Note that $OP\perp m$. Let $EF$ be the diameter of $(A)$ perpendicular to $m$. Point $L$ of tangency of the two circles serves as their center of similitude so that, besides lying on $AO$ it also belongs to (in the diagram) $EP.$ Since $EF$ is a diameter, $\angle ELF$ is right, as is $\angle EGP$, making quadrilateral $FLPG$ cyclic. By the Intersecting Secants Theorem,

$EF\cdot EG = EL\cdot EP = EB\cdot EC$

where $C$ is the second point of intersection of $EB$ with $(O)$. By the inverse of the Intersecting Secants Theorem, quadrilateral $FCBG$ is cyclic, such that $C$ can also be determined as the intersection of $EB$ and $(FCBG)$. Thus circles that solve the problem pass through two point $B$ and $C$ and are tangent to either $m$ or $(A)$. The PLC problem is thus reduced to a choice of either PPL or PPC.

The construction starts with finding $C$ and continuing with one of the PPL or PPC constructions, does not matter which. Either leads to two solutions. Swapping the roles of $E$ and $F$ leads to two more solutions, making the total number of solutions 4.

The construction does not work when $m$ separates $B$ and $C$, or when $B$ lies on $EF,$ although, in the latter case, two solutions may exist when $G\not\in EF$ while $B\in FG.$

### References

1. N. Altshiller-Court, College Geometry, Dover, 1980, #512