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Mathematics
Mathematics
Problem
Prove that if the four points of intersection of the circles
x
2
+
y
2
+
a
x
+
b
y
+
c
=
0
x^2 + y^2 + ax + by + c = 0
x
2
+
y
2
+
a
x
+
b
y
+
c
=
0
and
x
2
+
y
2
+
a
′
x
+
b
′
y
+
c
′
=
0
x^2 + y^2 + a'x + b'y + c' = 0
x
2
+
y
2
+
a
′
x
+
b
′
y
+
c
′
=
0
by the lines
A
x
+
B
y
+
C
=
0
Ax + By + C = 0
A
x
+
B
y
+
C
=
0
and
A
′
x
+
B
′
y
+
C
′
=
0
A'x + B'y + C' = 0
A
′
x
+
B
′
y
+
C
′
=
0
respectively are concyclic, then
∣
a
−
a
′
b
−
b
′
c
−
c
′
A
B
C
A
′
B
′
C
′
∣
=
0.
\begin{vmatrix} a-a' & b-b' & c-c' \\ A & B & C \\ A' & B' & C' \end{vmatrix} = 0.
a
−
a
′
A
A
′
b
−
b
′
B
B
′
c
−
c
′
C
C
′
=
0.
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