(x-x0)^2+(y-y0)^2=0 canonical form

Given equation of the surface of 2-order:
$$\left(x - x_{0}\right)^{2} + \left(y - y_{0}\right)^{2} = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x y_{0} + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y y_{0} + 2 a_{24} y + a_{33} y_{0}^{2} + 2 a_{34} y_{0} + a_{44} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = - x_{0}$$
$$a_{22} = 1$$
$$a_{23} = -1$$
$$a_{24} = 0$$
$$a_{33} = 1$$
$$a_{34} = 0$$
$$a_{44} = x_{0}^{2}$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22} + a_{33}$$

     |a11  a12|   |a22  a23|   |a11  a13|
I2 = |        | + |        | + |        |
     |a12  a22|   |a23  a33|   |a13  a33|

$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|$$

     |a11  a14|   |a22  a24|   |a33  a34|
K2 = |        | + |        | + |        |
     |a14  a44|   |a24  a44|   |a34  a44|

     |a11  a12  a14|   |a22  a23  a24|   |a11  a13  a14|
     |             |   |             |   |             |
K3 = |a12  a22  a24| + |a23  a33  a34| + |a13  a33  a34|
     |             |   |             |   |             |
     |a14  a24  a44|   |a24  a34  a44|   |a14  a34  a44|

substitute coefficients
$$I_{1} = 3$$

     |1  0|   |1   -1|   |1  0|
I2 = |    | + |      | + |    |
     |0  1|   |-1  1 |   |0  1|

$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & 1 & -1\\0 & -1 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 0 & 0 & - x_{0}\\0 & 1 & -1 & 0\\0 & -1 & 1 & 0\\- x_{0} & 0 & 0 & x_{0}^{2}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0 & 0\\0 & 1 - \lambda & -1\\0 & -1 & 1 - \lambda\end{matrix}\right|$$

     | 1   -x0|   |1   0 |   |1   0 |
     |        |   |      |   |      |
K2 = |       2| + |     2| + |     2|
     |-x0  x0 |   |0  x0 |   |0  x0 |
           

     | 1   0  -x0|   |1   -1   0 |   | 1   0  -x0|
     |           |   |           |   |           |
     | 0   1   0 |   |-1  1    0 |   | 0   1   0 |
K3 = |           | + |           | + |           |
     |          2|   |          2|   |          2|
     |-x0  0  x0 |   |0   0   x0 |   |-x0  0  x0 |
           

$$I_{1} = 3$$
$$I_{2} = 2$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 3 \lambda^{2} - 2 \lambda$$
$$K_{2} = 2 x_{0}^{2}$$
$$K_{3} = 0$$
Because
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
then by type of surface:
you need to
Make the characteristic equation for the surface:
$$- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0$$
or
$$\lambda^{3} - 3 \lambda^{2} + 2 \lambda = 0$$
Solve this equation
$$\lambda_{1} = 2$$
$$\lambda_{2} = 1$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$2 \tilde x^{2} + \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{2}}{2}\right)^{2}} + \frac{\tilde y^{2}}{1^{2}} = 0$$
this equation is fora type two imaginary intersecting planes
- reduced to canonical form