by John Moore and Marc Lewis

If we are given an arbitrary line, **l**, with an arbitrary point,
**p**, and an arbitrary circle, **O**, (with center **k**) not
intersecting the line or point, then we would like to find two circles which
are tangent to our given point and circle. To begin this, let's construct a
line perpendicular to line **l** through **p**. Call it **m**. One
thing that we know is that the centers for both of our desired circles will
fall on this line.

Let's take the radius of circle **O** (labeled **r**) and construct a
circle, **Q**, of radius **r** with center **p**. Label the points of
circle **Q** that intersect line **m**, **a** and **b**. Let's now
draw a segment connecting **k** and **a**. Construct the perpendicular
bisector of this segment. Where this perpendicular bisector intersects line
**m** (labeled **c**)** **is the center of the first of our desired
circles. This circle should intersect point **p**.

A similar process is performed to find the other circle. However this time,
draw a segment connecting **k** and **b**. The rest of the construction
is the same. Now let's have a look at this construction...

Let's call the center of the first found circle **f** and the center of the
other circle **g**. I want to deal with the first circle, appropriately,
first. Draw a segment from **f** to **k**. Now, I want to claim that
**af** is congruent to **kf**. We know that we found **f** by
constructing the perpendicular bisector of **ak**. Let's call this point of
bisection **c**. Therefore, angles **acf** and **kcf** are right
angles and are congruent (since all right angles are congruent and by the
definition of perpendicular). We also know that segment **cf** is congruent
to segment **cf** because every segment is congruent to itself. Also, we
have that segments **ac** and **kc** are congruent, by the definition of
bisector. Therefore, triangles **acf** and **kcf** are congruent by the
side-angle-side postulate. So, segments **af** and **kf** are congrent
because the corresponding parts of congruent triangles are congruent.

Here's a picture to let you see the letters...

A similar proof for our circle with center **g** will show that segments
**gb** and **gk** are congruent. Now, if we construct a line **w**
parallel to line **l** through point **a**, **w** should be the
directrix and point **k** should be the focus for a parabola opening to the
right with its vertex at point **f** when it is collinear with points
**a**, **p**, and **k**. Similarly, if we construct a line **x**
parallel to line **l** through point **b**, **x** should be the
directrix and **k** should be the focus for a parabola opening to the right
with its vertex at point **g** when it is collinear with points **b**,
**p**, and **k**. Since we've already proven for arbitrary objects that
the distance between the parabola and these points are equal, a GSP sketch
should nail it down for those who do not believe.