Difference between revisions of "Radical axis"
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If a line drawn through point P intersects circle <math>\omega</math> at points A and B, then <math>pow(P, \omega) = PA * PB</math>. | If a line drawn through point P intersects circle <math>\omega</math> at points A and B, then <math>pow(P, \omega) = PA * PB</math>. | ||
'''Theorem 2: (Radical Axis Theorem)''' | '''Theorem 2: (Radical Axis Theorem)''' | ||
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a. The radical axis is a line perpendicular to the line connecting the circles' centers (line <math>l</math>). | a. The radical axis is a line perpendicular to the line connecting the circles' centers (line <math>l</math>). | ||
Revision as of 18:04, 4 June 2014
Work in Progress
Definitions
The power of point 
with respect to circle 
(with radius 
and center 
, which shall thereafter be dubbed 
, is defined to equal 
.
The radical axis of two circles 
is defined as the locus of the points 
such that the power of with respect to 
and 
are equal. In other words, if 
are the center and radius of 
, then a point is on the radical axis if and only if ![\[PO_1^2 - r_1^2 = PO_2^2 - r_2^2\]](http://latex.artofproblemsolving.com/8/7/b/87be0c12637822df7a8654ea24005bf39ae11175.png)
Results
Theorem 1: (Power of a Point)
If a line drawn through point P intersects circle at points A and B, then 
.
Theorem 2: (Radical Axis Theorem)
a. The radical axis is a line perpendicular to the line connecting the circles' centers (line 
).
b. If the two circles intersect at two common points, their radical axis is the line through these two points.
c. If they intersect at one point, their radical axis is the common internal tangent.
d. If the circles do not intersect, their radical axis is the perpendicular to through point A, the point on such that 
.
Theorem 3: (Radical Axis Concurrence Theorem) The three pairwise radical axes of three circles concur at a point, called the radical center.
Proofs
Theorem 1 is trivial Power of a Point, and thus is left to the reader as an exercise. (Hint: Draw a line through P and the center.)
Theorem 2 shall be proved here.
