Article 34
Some Circles in the Cyclic Quadrilateral configuration
Christopher J Bradley
E
Q
A
S
D
O
G
B
T
P
R
C
F
Figure
1. Introduction
When ABCD is a cyclic quadrilateral, centre O, and BA^CD = E, BC^AD = F and AC^BD
= G, then circles ABF, CDF, BCE, ADE meet at a point T on EF. The perpendicular from
T to EF passes through G and O. The centres of the four circles, O and T lie on a circle.
These facts are established using Cartesian co-ordinates. These are not new results, though
until now I did not know that the last mentioned circle passes through T as well as the five
circle centres. See the figure above.
2. Preliminary analysis and the diagonal points E, F, G, and T
We take ABCD to be the unit circle, origin the centre O, and A with co-ordinates (2a/(1 +
a2), (1 – a2)/(1 + a2)). B, C, D have similar co-ordinates but with parameters b, c, d instead
of a. The chord AB has equation
(a + b)x + (1 – ab)y = (1 + ab),
(2.1)
Other chords have similar equations with appropriate change of parameters.
The chords AB and CD meet at E with co-ordinates (x, y), where
x = 2(ab – cd)/(abc + abd – acd – bcd + a + b – c – d),
y = – (abc + abd – acd – bcd – a – b + c + d)/( abc + abd – acd – bcd + a + b – c – d). (2.2)
F = BC^AD has similar co-ordinates to those in Equation (2.1), but with a and c
interchanged. G = AC^BD has similar co-ordinates to those in Equation (2.1), but with b
and c interchanged.
The equation of EF is found to be
2(ac – bd)x – (abc – abd +acd – bcd – a + b – c +d)y
– abc + abd – acd + bcd – a + b – c + d) = 0.
(2.3)
The perpendicular to EF through O has equation
2(ac – bd)y + (abc – abd +acd – bcd – a + b – c +d)x = 0.
(2.4)
It may now be checked that this line passes through G. Indeed it is well known that O is
the orthocentre of triangle EFG.
The lines with Equations (2.3) and (2.4) meet at a point T with co-ordinates (m/k, n/k),
where
m = 2(a2c(b(c – d) + cd + 1) – a(b2d(c – d) + b(c2d + c(d2 + 1) + d) – c(c – d)) + bd(b(cd +
1) – c + d)),
n = – (a(b(c – d) + cd + 1) – b(cd + 1) + c – d)(a(b(c – d) + cd – 1) + b(1 – cd) – c + d)(2.5)
k = a2(b2(c – d)2 + 2b(c – d)(cd – 1) + c2(d2 + 4) – 2cd + 1) – 2a(b2(c – d)(cd – 1) + b(c2 +
1)(d2 + 1) + (c – d)(cd – 1)) + b2(c2d2 – 2cd + 4d2 + 1) + 2b(c – d) (cd – 1) + (c – d)2.
3. The four circles through T
Every circle may be expressed in the form
x2 + y2 + 2gx + 2fy + k = 0,
and since this may be written in the form
(x + g)2 + (y + f)2 = g2 + f2 – k,
(3.1)
(3.2)
it represents a circle, centre (– g, – f) and radius √(g2 + f2 – k).
In order to obtain the equation of a circle through three given points one can put the coordinates of those points in turn into Equation (3.1), solve the three resulting equations for
f, g, k and insert those values back into the equation. To specify a circle one may write
down its equation or specify its values of f, g, k. The latter is useful if the centre is what is
required.
The equation of circle ABF turns out to be
(abc + bcd – abd – acd – a + b + c – d)(x2 + y2) + 2(a + b)(d – c)x + 2(c – d)(ab – 1)y
+ abc – abd + acd – bcd + a – b + c – d = 0.
(3.3)
It may now be verified that this circle passes through the point T.
Similarly the circles CDF, BCE, ADE all pass through T.
4.
The six point circle PQRSOT
The points P, Q, R, S are the centres of circles ABF, BCE, CDF, ADE respectively. Their
co-ordinates are:
For P:
For Q:
((a + b)(c – d), (1 – ab)(c – d))/(abc – abd – acd + bcd – a + b + c – d);
((b + c)(a – d), (1 – bc)(a – d))/(abc + abd – acd – bcd + a + b – c – d);
(4.1)
(4.2)
For R:
((b – a)(c + d), (b – a)(1 – cd))/(abc – abd – acd + bcd – a + b + c – d);
(4.3)
For S:
((a + d)(b – c), (1 – ad)(b – c))/abc + abd – acd – bcd + a + b – c – d);
(4.4)
The equation of the circle that contains all these points and also O and T is
(abc + abd – acd – bcd + a + b – c – d)(abc – abd – acd + bcd – a + b +c – d)(x2 + y2)
+ (a2(b2(c – d)(cd – 1) + b(1 + c2)(1 + d2) +(c – d)(cd – 1)) – a(b2 + 1)(c2 + 1)(d2 + 1) +
b2(c – d)(cd – 1) + b(c2 + 1)(d2 + 1) + (c – d)(cd – 1))x + (a2(b2(c + d)(c – d) + c2(d2 + 2) +
1) – b2(c2d2 + 2d2 + 1) + c2 – d2)y = 0.
(4.5)
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