Let three circles be located such that their centers are non-collinear.
Construct a fourth circle that is
orthogonal to each of the three circles.
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Submitted by Bractals
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Rating: 4.0000 (1 votes)
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Solution:
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CONSTRUCTION:
Let circle C4 intersect circles C1, C2, and C3 at
points P41, Q41, P42, Q42, P43,
and Q43 respectively.
Let circle C5 intersect circles C1,
C2, and C3 at
points P51, Q51, P52,
Q52, P53, and Q53 respectively.
Let
X412 = P41Q41 /\ P42Q42
X512 = P51Q51 /\ P52Q52
X423 = P42Q42 /\ P43Q43
X523 = P52Q52 /\ P53Q53
R = X412X512 /\ X423X523
where /\ denotes intersection.
The desired circle is a circle with center R and orthogonal to any one of the circles C1, C2, and C3.
PROOF:
PxyQxy is the
radical axis of circles Cx and Cy.
Xxyz is a point on the radical axis of circles Cy and Cz.
X4yzX5yz is the radical axis of circles Cy and Cz.
R is the
radical center of circles C1, C2, and C3.
A circle with center R and orthogonal to any one of the
circles C1, C2, and C3
is orthogonal to all three and is called the
radical circle.
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