To deceased Professor: Mohsen Hashtroudi
Four Dimensional Space of Circles
Alireza Amir Moez1
Deceased Professor Hashtroudi generalized the inner product of two vectors to the inner
product of two circles and described its applications in geometry of circles and lines 2. This
interesting and useful subject was also translated into English3. Now, we briefly survey some
invariant properties of this product. Then we show that the set of circles and lines in a plane
form a 4-dimensional space in that the inner product of circles is its same vector product.
1. The scalar product of two circles: we suppose that
a ( x ² + y ²) − 2bx − 2cy + d = 0
(1)
is the equation of a circle with radius r. In this equation, if a = 0, the a direct line is
obtained. Hence, the equation (1) shows a circle or a line. Now, we consider two circles
C1 and C 2 with these equations respectively:
C1 : a1 ( x ² + y ²) − 2b1 x − 2c1 y + d 1 = 0
C 2 : a 2 ( x ² + y ²) − 2b2 x − 2c2 y + d 2 = 0
assume that the radii of C1 and C 2 are respectively r1 and r2 . Then we define the scalar or
inner product of C1 and C 2 as following:
(C1 , C 2 ) = ( a1r1 )( a 2 r2 ) cos α
in that α is the angle between two circles. We leave details to readers who refer to [2].
is directed angle. If we obtain the value of ( C1 , C 2 ) in terms of coefficents,
(C1 , C 2 ) = b1b2 + c1c2 −
a1d 2 + a 2 d 1
2
From this the inner product is obtained that the norm of the circle (1) is equal to:
1
|| C ||= (C , C ) 2 = b 2 + c 2 − ad = ar
indeed, the above value is pseudo norm because isotrop circles are created, namely acircle
which is not zero, its norm may be equal to zero; for instance:
x 2 + y 2 − 2 x − 4 y + 20 = 0
However, this circle is virtual. Reader can easily show that if a goes to zero, the value of r
will go to infinity and the value of ||C|| is again meaningful, namely:
|| C ||= b 2 + c 2
however, in this case the scalar product of two circles (two lines) changes to the scalar
product of two vestors which are prependicular to both lines.
2. Invariability of the product of two circles under transfer: the deceased Professor
Mc Duffee proved that the product of ( C1 , C 2 ) in part one remains constant under
transfer and he asked which other transformations hold constantly this product.
Proof: We assume that C1 and C 2 are the circles of part one and we consider the transfer
equations:
x=X+h
and y=Y+k
we see that after transfer, (1) in part one changes to:
C ' : a ( X ² + Y ²) − 2(b − ah ) X − 2( a − ak )Y − 2bk − 2ck + ah ² + ak ² + d = 0
If we consider the value of C1 and C 2 and we compute ( C1 ´, C 2 ´) , it is again equal to:
b1b2 + c1c2 −
a1d 2 + a 2 d 1
2
3. Invariability of the product of circles under rotation: we consider again the circles
C1 and C 2 in part one. The equations of rotation are generally in the plane:
x = hX − kY
, h² + k ² = 1
y = kX + hY
Easily we can see that after rotation, the equation C in part one is:
C ' : a ( X 2 + Y 2 ) − 2(bh + ck ) X − 2( ch − bk )Y + d = 0
We assume again that C1 ´ and C 2 ´ are transformer of C1 and C 2 . We see that
(C1 ' , C 2 ' ) = (b1h + c1k )(b2 h − c2 k ) + ( c1h − b1k )( c2 h − b2 k ) −
= b1b2 + c1c2 −
4. Reflection: we consider the circle C in part one and the following reflection:
a1d 2 + a 2 d 1
2
a1d 2 + a 2 d 1
2
kX
x= 2
X +Y2
y = kY
X² +Y²
( X , Y ) ≠ (0,0)
and
We see that the transformer of C becomes:
C ' : d ( X ² + Y ²) − 2bkX − 2ckY + ak ² = 0
(C1 ' , C 2 ' ) = b1b2 k ² + c1c2 k 2 −
( a1d 2 + a 2 d 1 )k 2
2
±
1, then ( C1 , C 2 ) will remain constant.
if k =
Still this question remains: in general under which transformations does the value of ( C1 ,
C 2 ) remain constant? Are these transformations confined to the parts 2, 3, 4?
5. 4-dimensional space of circles: we consider again the circle C in part one, we see that
the C can be written as:
a+d
a−d
[ x ² + y ² + 1] − 2bx − 2cy +
2
2
C:
we consider the following elements as base:
B1 = x ² + y ² + 1 = 0 , B2 = −2 x = 0 , B3 = −2 y = 0 , B4 = x ² + y ² − 1 = 0
we see that C is a linear combination of B1 , B2 , B3 , B4 and B1 is a complex circle. Reader
can easily prove that the set of circles forms a vector space on real numbers. If we
consider C in the following form:
C: pB1 + bB2 + cB3 + qB4
The product ( C1 , C 2 ) will be equal to:
(C1 , C 2 ) = − p1 p2 + b1b2 + c1c2 + q1q2
that its difference with common vector product is just the sign (-) in the first sentence. We
leave it to readers.
6. Transformations on the space of circles: among transformations in a vector space,
transfer and reflection are not linear. But if we apply the space of circles, these
transformations will become linear. At first, we survey transfer. We assume that:
F(x, y) = (x + h, y + k) = (X, Y), x = X – h , y = Y - k
Therefore, we consider the circle C in part one:
a+d
a−d
B1 + bB2 + cB3 +
B4 = 0
2
C: 2
Since C is a linear combination of B1 , B2 , B3 , B4 , its transfer will be a linear combination
h² + k ² + 2
h² + k ²
B1 + hB2 + kB3 +
2
2
F ( B2 ) = −hB1 + B2 + hB4
F ( B1 ) =
F ( B3 ) = −kB1 + B3 + kB4
h2 + k 2
2 − h² − k ²
B1 + hB2 + kB3 +
2
2
F ( B4 ) =
So, the transformation has a matrix which is as follows:
h² + k ² + 2
2
−h
−k
h² + k ²
2
h k
1 0
0 1
h k
h² + k ²
2
h
k
2 − h² − k ²
2
Now, we survey the reflection:
kX
x = X ² + Y ²
y = kY
X² +Y²
G:
1+ k²
1− k²
B1 +
B4
2
2
G ( B2 ) = B2
G ( B1 ) =
G ( B3 ) = B3
G ( B4 ) =
1− k²
1+ k²
B1 +
B4
2
2
so the reflection matrix will be as follows:
1 + k ²
2
0
0
1 − k ²
2
1− k²
2
1 0
0
0 1
0
1+ k²
0 0
2
0 0
We leave the other transformations to readers.
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