**Exploration 7: **

**Tangent Circles**

John Watson

Part I: Given two circles and a point on one of the circles. Construct two different circles tangent to the two given circles with one point of tangency being the designated point.

Lets start by constructing two circles.

Circle A will be centered at point A.

Circle B will have its center at point B.

Construct an arbitrary point X on Circle A.

Point X will be our point of tangency.

How would we go about constructing a new circle that passes through point X and is tangent to Circle A and Circle B?

If the circle passes through point X and is tangent to Circle A, then its center must be on the line that passes through line XA. Let’s construct line XA…

(We know both of our tangent circles will be centered on this line.)

Let’s construct a circle centered at point X with the same length radius as circle B.

Notice that…

Length of segment X’A = Length of radius A + Length of radius B

Now lets construct a line segment from B to X’ and label its midpoint M.

.Then construct a perpendicular bisector to segment BX’ and label the intersection with line AX’ as point C.

Notice how triangle CBX’ is an isosceles triangle, which means that…

Length of segment CX’ =Length of segment CB

.Point C will be the center of our first tangent circle.

Construct Circle C with its center at point C and a radius of length CX.

Notice how Circle C is tangent to Circle A and Circle B.

.We will construct our second tangent circle in a similar way.

Let’s construct a circle centered at point X with the same length radius as circle B.

This time, notice that...

Length of segment AX’’ = Length of segment AX - Length of segment XX’’

Now lets construct a line segment from B to X’’ and label its midpoint M’.

Then construct a perpendicular bisector to segment BX’’ and label the intersection with line AX’’ as point D.

Notice how triangle DBX’’ is an isosceles triangle, which means that…

Length of segment DX’’ = Length of segment DB

.

Point D will be the center of our other tangent circle.

Construct Circle D with its center at point D and a radius of length DX.

Notice how Circle D is also tangent to Circle A and Circle B.

Here is a **script tool** for these types of constructions…

*Open the script view for explicit instructions on how to use the tool.*

Now let's take a look at the traces of the centers of our tangent circles as the tangent point X changes position.

In the example below, (as in the previous example), one of the given circles is completely inside the other given circle.

Click **here** to see traces of the centers of the tangent circles (Point C and Point D) as the tangent Point X changes position on the given Circle A.

Notice how the centers of the given circles are foci for both of the ellipses generated by these traces.

Here is a construction where the two given circles over lap...

Click** here** to see traces of the centers of the tangent circles (Point C and Point D) as the tangent Point X changes position on the given Circle A.

Notice how the centers of the given circles are foci for the of the ellipse generated by the trace of Point C.

Notice how the centers of the given circles are foci for the hyperbola generated by the trace of Point D.

Also, notice how the tangent circles collapse to a single point when their centers are located at the intersections of the given circles.

Here is a construction where the two given circles are disjoint...

Click

hereto see traces of the centers of the tangent circles (Point C and Point D) as the tangent Point X changes position on the given Circle A.Notice how the centers of the given circles are foci for both of the hyperbolae generated by these traces.

Part II:Given a circleand a line, construct a circle that is tangent to the given circle and the given line.

Let’s choose point X to be our point of tangency on our given circle M. As in the previous exercise, the only way for our tangent circle to pass through point X and be tangent to Circle M is if the center of the tangent circle lies on the line MX. Let’s construct line MX. Label the intersection of line MX and our given line as point Y.

Construct a line that is perpendicular to line MX and passes through point X. Label the intersection with our given line as point Z. Notice how line XZ is tangent to the given Circle M. Also, note that triangle XYZ is a right triangle.

We know the tangent circle will pass through point X and one point on our given line. Since all radii of a circle are the same length, the center of our tangent circle will be equidistant from point X and our given line. Construct the angle bisector of angle XZY. (This line represents all of the points that are equidistant from line XZ and the given line.) Label the intersection of the angle bisector and line MX as point N.

We know point N is located on line MX and is equidistant from point X and the given line. So, point N is the center of our tangent circle.

Hereis an animation of the construction as point X moves around the given circle M.Recall that angle YXZ is a right angle, which makes angle XYZ and angle XZY complimentary angles. Notice how triangle XYZ "jumps" to the other side of line XZ whenever one of the complimentary angles reaches 90 degrees. This happens because the three angles of a triangle always add up to 180 degrees.

Hereis a link to a script tool for constructing a tangent circle given a line and a circle.

Open the script view for explicit instructions on how to use the tool.

.

.