SECTION 1.3 Circle Geometry 31
- Let 4 ABChave circumradiusR. Show that
Area 4 ABC=R(acosA+bcosB+ccosB)
2,
wherea=BC, b=AC, andc=AB. (See exercise 5, page 17 for
the corresponding result for the inscribed circle.)Circle of Apollonius
Circle of Apollonius. Assume that c 6 = 1is a constant and that
AandBare two given points. Then the locus of points
{
P∣∣
∣PA
PB
=c}is a circle.
Proof. This is actually a very sim-
ple application of the Angle Bisec-
tor Theorem (see also Exercise 1,
page 16). LetP 1 andP 2 lie on the
line (AB) subject to
AP 1
P 1 B=c=AP 2
BP 2
.
If we letP an arbitrary point also subject to the same condition, then
from the Angle Bisector Theorem we infer that APP̂ 1 = P 1 PB̂ and
BPP̂ 2 = 180−APB.̂
This instantly implies thatP 1 PP̂ 2 is a right angle, from which we con-
clude (from Exercise 2, page 30 above) thatP sits on the circle with
diameter [P 1 P 2 ], proving the result.