Excercise 40 Prove that the pairwise radical axes of three circles with not collinear centers meet a сотmon point. Hint. Equalities O,p² – rỉ = 02P² – rž and O2P² – r = O3P2 – r? - imply that O1P² – rỉ = O3P² – rž, i.e. the intersection point of the radical axes l12 and l23 is on l13 as well. It is called the radical center of the circles.

Elementary Geometry for College Students
6th Edition
ISBN:9781285195698
Author:Daniel C. Alexander, Geralyn M. Koeberlein
Publisher:Daniel C. Alexander, Geralyn M. Koeberlein
Chapter10: Analytic Geometry
Section10.4: Analytic Proofs
Problem 28E
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Excercise 40 Prove that the pairwise radical axes of three circles with not collinear centers meet a
сотmоn pоint.
Hint. Equalities
0,p² – r} = O2P2 – rž and O,P2 – r = O3P? – r
imply that O1P² – r? = O3P² – r, i.e. the intersection point of the radical axes l12 and l23 is on l13
as well. It is called the radical center of the circles.
Transcribed Image Text:Excercise 40 Prove that the pairwise radical axes of three circles with not collinear centers meet a сотmоn pоint. Hint. Equalities 0,p² – r} = O2P2 – rž and O,P2 – r = O3P? – r imply that O1P² – r? = O3P² – r, i.e. the intersection point of the radical axes l12 and l23 is on l13 as well. It is called the radical center of the circles.
Expert Solution
Step 1

Let A,B and C be three circles with no collinear centers.

Let r1 be the intersection point and l12 be the radical axis of circles A and B.

Let r2 be the intersection point and l23 be the radical axis of circles B and C

Let r3 be the intersection point and l31 be the radical axis of circles C and A.

Let O1, O2, O3 be the centers of the three circles.

Let P be a point, then the radical axis of circles A and B is defined as the line along which the tangents to those circles are equal in length.

O1P2-r12=O2P2-r22

Similarly, the tangents to circles B and C must be equal in length on their radical axis.

O2P2-r22=O3P2-r32

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