When you define a **beam circumferences** infinity as a set simply fulfilling a restriction based on the **power**, sorted the beams depending on the relative position of its elements.

Los **parabolic beams circumferences** These families are among circumferences. We will see how to determine elements that belong.

Given the circles tangent to a point **O**, the **radical axis “and” circumferences** coincides with the common tangent to two circles. This line is perpendicular to the one containing the centers of the circumferences.

The endless circles tangent two circles tangent to each other at a point

O,determine oneparabolic beam circumferences. Thepunto Ocalledbeam center.

The radical axis of any two circles of this bundle is the line **and**.

All centers of the circumferences of the bundle are in a straight,

b, denominadastraight beam foundation.

### Determine a circumference of the parabolic beam passing through a point P

From the endless circles of parabolic beam, only passes through a given point than the center **O** beam. Let's see how to determine the center of a circle of the beam passing through a point **P** any.

The circumference sought O1 will be centered in the base line, b, and pass through the points P and O, so shall also the bisector of these points.

Solution, its center, thus determined by the intersection of two loci, the base line and the bisector of the PO segment containing two crossing points.

### Determine the parabolic beam circumferences are tangent to a given line

Tangency condition is determined by a straight **t** anyone who does not match with right or basis **b** or the radical axis **and**.

To solve the problem look for a point **Cr**, radical shaft **and**, have equal power with respect to the beam circumferences, and belonging, and the embroidery, to the line **t ya **the latter is the center of the circumferences radical which are tangent. We see, that **Cr** is the radical center of the line **t** (infinite radius circumference) and parabolic beam circumferences.

As shown in Figure, power **Cr** on all beam circumferences finding can determine the distance (squared) center **O** beam. This distance is to be also the points of tangency of the solutions sought. We have two solutions because we can take this away **Cr-O** on both sides of **Cr** on the line **t**.

### Determine the parabolic beam circumferences are tangent to a given circle

The generalization of the problem comes when the condition of tangency is on a circle t any.

In this case, again, determine a point **Cr** have equal power with respect to the circumference marking the tangency condition and any of the parabolic beam, so it must be in its radical axis.

The solutions pass through the points **T1** and **T2** located on tangents drawn from **Cr**, and which are at the root of the distance power that we have calculated as in the previous case.

The centers of the solutions found aligned with the center of the circle **t** and corresponding contact points.

## Make conjugate

Last, we can see in the figure below the conjugate beam (orthogonal) of a parabolic beam, it can be deduced that is another parabolic base line the radical axis of the front.