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Canonical form/ x^2+y^2-1

x^2+y^2-1 canonical form

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The solution

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      2    2    
-1 + x  + y  = 0
x2+y2−1=0x^{2} + y^{2} - 1 = 0 
x^2 + y^2 - 1 = 0
Detail solution
Given line equation of 2-order:
x2+y2−1=0x^{2} + y^{2} - 1 = 0
This equation looks like:
a11x2+2a12xy+2a13x+a22y2+2a23y+a33=0a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0
where
a11=1a_{11} = 1
a12=0a_{12} = 0
a13=0a_{13} = 0
a22=1a_{22} = 1
a23=0a_{23} = 0
a33=−1a_{33} = -1
To calculate the determinant
Δ=∣a11a12a12a22∣\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|
or, substitute
Δ=∣1001∣\Delta = \left|\begin{matrix}1 & 0\\0 & 1\end{matrix}\right|
Δ=1\Delta = 1
Because
Δ\Delta
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
a11x0+a12y0+a13=0a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0
a12x0+a22y0+a23=0a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0
substitute coefficients
x0=0x_{0} = 0
y0=0y_{0} = 0
then
x0=0x_{0} = 0
y0=0y_{0} = 0
Thus, we have the equation in the coordinate system O'x'y'
a33′+a11x′2+2a12x′y′+a22y′2=0a'_{33} + a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} = 0
where
a33′=a13x0+a23y0+a33a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}
or
a33′=−1a'_{33} = -1
a33′=−1a'_{33} = -1
then equation turns into
x′2+y′2−1=0x'^{2} + y'^{2} - 1 = 0
Given equation is circle
x~2(1−1)2+y~2(1−1)2=1\frac{\tilde x^{2}}{\left(1^{-1}\right)^{2}} + \frac{\tilde y^{2}}{\left(1^{-1}\right)^{2}} = 1
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)

Basis of the canonical coordinate system
e⃗1=(1, 0)\vec e_1 = \left( 1, \ 0\right)
e⃗2=(0, 1)\vec e_2 = \left( 0, \ 1\right)
Invariants method
Given line equation of 2-order:
x2+y2−1=0x^{2} + y^{2} - 1 = 0
This equation looks like:
a11x2+2a12xy+2a13x+a22y2+2a23y+a33=0a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0
where
a11=1a_{11} = 1
a12=0a_{12} = 0
a13=0a_{13} = 0
a22=1a_{22} = 1
a23=0a_{23} = 0
a33=−1a_{33} = -1
The invariants of the equation when converting coordinates are determinants:
I1=a11+a22I_{1} = a_{11} + a_{22}
     |a11  a12|
I2 = |        |
     |a12  a22|

I3=∣a11a12a13a12a22a23a13a23a33∣I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|
I(λ)=∣a11−λa12a12a22−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|
     |a11  a13|   |a22  a23|
K2 = |        | + |        |
     |a13  a33|   |a23  a33|

substitute coefficients
I1=2I_{1} = 2
     |1  0|
I2 = |    |
     |0  1|

I3=∣10001000−1∣I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & -1\end{matrix}\right|
I(λ)=∣1−λ001−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & 1 - \lambda\end{matrix}\right|
     |1  0 |   |1  0 |
K2 = |     | + |     |
     |0  -1|   |0  -1|

I1=2I_{1} = 2
I2=1I_{2} = 1
I3=−1I_{3} = -1
I(λ)=λ2−2λ+1I{\left(\lambda \right)} = \lambda^{2} - 2 \lambda + 1
K2=−2K_{2} = -2
Because
I2>0∧I1I3<0I_{2} > 0 \wedge I_{1} I_{3} < 0
then by line type:
this equation is of type : circle
Make the characteristic equation for the line:
−I1λ+I2+λ2=0- I_{1} \lambda + I_{2} + \lambda^{2} = 0
or
λ2−2λ+1=0\lambda^{2} - 2 \lambda + 1 = 0
λ1=1\lambda_{1} = 1
λ2=1\lambda_{2} = 1
then the canonical form of the equation will be
x~2λ1+y~2λ2+I3I2=0\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2} + \frac{I_{3}}{I_{2}} = 0
or
x~2+y~2−1=0\tilde x^{2} + \tilde y^{2} - 1 = 0
x~2(1−1)2+y~2(1−1)2=1\frac{\tilde x^{2}}{\left(1^{-1}\right)^{2}} + \frac{\tilde y^{2}}{\left(1^{-1}\right)^{2}} = 1
- reduced to canonical form
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