## Wednesday, December 9, 2015

### Unizor - Geometry2D - Apollonius Problems - Lines and Circles

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

The problems below are easily solvable using the transformation of inversion (symmetry relative to a circle). We recommend to review a previous lecture dedicated to this topic.

The main technique we will use is transformation of a circle passing through a center of an inversion circle into a straight line.

Here is a quick reminder of how inversion works.

An inversion circle with radius R and center O is given. Then any point P (other than center O) is transformed into point P' such that P' lies on a ray from center O to given point P and OP·OP'=R²

We have proven that any circle passing through a center of an inversion circle is transformed by an inversion into a straight line and a circle not passing through a center of inversion is transformed into a circle.

Apollonius Problems -

Lines and Circles

Problem LLC

Construct a circle tangential to two given lines, a and b, and tangential to a given circle c, presuming that both lines are located outside of circle c.

Solution

Analysis:

Assume, our circle is constructed. It is tangential to lines a and b and to circle c.

We can easily reduce this problem to the one we solved before - construction of a circle tangential to two lines and passing through a point.

To accomplish this, increase the radius of a presumably constructed circle by the radius of a given circle and shift both given lines, a and b, away from a given circle in between them by the same value to new positions - a' and b'.

Now a new circle, concentric with the one we have to construct and having a radius greater than that by a radius of a given circle c, will be tangential to lines a' and b' and passing through a center of a given circle c.

Construction:

From a center of circle c drop a perpendicular onto line a and extend it by the radius of circle c beyond its base on line a to point A. Draw line a' parallel to line a through point A.

From a center of circle c drop a perpendicular onto line b and extend it by the radius of circle c beyond its base on line b to point B. Draw line b' parallel to line b through point B.

Construct a circle tangential to lines a' and b' and passing through a center of c.

Decrease the radius of this new circle by a radius of circle c leaving a center in place. This is a required circle.

Problem LCC

Construct a circle tangential to a given line a and tangential to two given circles, c and d, presuming that none of the given elements lies inside the other.

Solution

Analysis:

Assume, our circle is constructed. It is tangential to line a and to circles c and d.

Increase the radius of a presumably constructed circle by the smaller radius among two given circles (let's say, it's circle c), shift line a away from circle c parallel to itself by this same radius to a new positions a', and concentrically shrink circle d by the same radius to a new circle d'.

Now a new circle, concentric with the one we have to construct and having a radius greater than that by a radius of a given circle c, will be tangential to line a', to circle d' and passing through a center of a given circle c.

We can construct it by introducing an inversion with a center at the center of circle c that transforms line a' into some circle aa, circle d' - also into some circle dd, and the circle concentric to the one we need to construct - into a line tangential to these two circles. aa and dd.

Construction:

From a center of circle c drop a perpendicular onto line a and extend it by the radius of circle c beyond its base on line a to point A. Draw line a' parallel to line a through point A.

Draw a circle d' concentric to circle d with a radius smaller than original by a radius of circle c.

Draw an inversion circle with a center at the center of circle c and invert relative to it line a' into circle aa and circle d' into circle dd.

Construct a line tangential to circles aa and dd.

Invert this line into a circle.

Decrease the radius of this new circle by a radius of circle c leaving a center in place. This is a required circle.

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