**Tangent Circle Construction**

**By Mary
Negley**

For
this assignment, I constructed tangent circles in a variety of situations. For my first case, I looked at the
following problem:

Given
a line and a circle with center *K*.
Take an arbitrary point *P* on the circle.
Construct two circles tangent to the given circle at *P* and tangent to the line.

In order to find the solution, I needed to think
through the properties of tangent circles. Below is a sketch of the line and circle with center *K*.

We
know that the circle, which is tangent to the circle with center *K*, will have its point of tangency with the circle at *P* because that is given. So the first step is to create a right angle at that
point. We do this by extending the
radius of the circle into a line at *P*
and creating a line perpendicular to the line through *K *and* P* at *P*.

To
construct a tangent circle, I bisected the angle *ABP *and I found the intersection of the angle bisector and
the line through *P* and *K*. I
labeled this intersection point *C*.

Then I constructed a circle centered at *C* with the radius *CP*.

Then, to construct the other tangent circle, I
bisected the angle *PBD* and found
the intersection of the angle bisector and the line through *P* and *K*. I labeled the intersection point *E*.

I then constructed a circle centered at *E* with the radius of *BE*.

Thus, I have constructed two circles tangent to the
given circle at *P* and tangent to
the line.

Suppose I wanted to construct a circle tangent to a
larger circle with a smaller circle inside it. See below for a possible sketch.

Using the strategy that I implemented above, I created
a line through *A *and *B *and found the intersections of the line with the
circles and labeled those.

An easy tangent circle to construct is the one tangent
to the larger circle at *C* and
tangent to the smaller circle at *D*. To construct that tangent circle, I
find the midpoint of the segment *CD*
and create a circle centered at that midpoint with the radius .

Another tangent circle may be formed by finding the
other point of intersection of the line and the smaller circle, labeling this
point *E *and finding the midpoint
of *CE* and creating a circle
centered at the midpoint with the radius.

A script of those two constructions may be found here.