Centers of Central Conic Section Curves

1. Real/Imaginary Ellipses, Real/Imaginary Circles and Hyperbolas are called Central Conic Section Curves as they have a Geometrical Center Point across which the Curves are Symmetric, i.e Any Line passing through this Point divides the Curves in 2 Equal Halves.
2. The Implicit Equation for any Central Conic Section Curve having its Center at the Origin is given as
   
   \(Ax^2 + Bxy + Cy^2 + F = 0\)   ...(1)
   
   On Translating the Center of the Conic Section given in equation (1) to a Point (\(x_c,y_c\)) (which is same as Translating the Equation of the Conic Section by (\(x_c,y_c\))), the Implicit Equation of the Conic Section Curve gets updated as
   
   \(A{(x-x_c)}^2 + B(x-x_c)(y-y_c) + C{(y-y_c)}^2 + F = 0\)
   
   \(\Rightarrow A(x^2 + {x_c}^2 - 2x_cx) + B(xy - y_cx - x_cy + x_cy_c) + C(y^2 + {y_c}^2 - 2y_cy) + F = 0\)
   
   \(\Rightarrow Ax^2 + A{x_c}^2 - 2Ax_cx + Bxy - By_cx - Bx_cy + Bx_cy_c + Cy^2 + C{y_c}^2 - 2Cy_cy + F = 0\)
   
   \(\Rightarrow Ax^2 + Bxy + Cy^2 + (- 2Ax_c - By_c)x + (- 2Cy_c - Bx_c)y + A{x_c}^2 + Bx_cy_c + C{y_c}^2 + F = 0\)   ...(2)
   
   \(\Rightarrow Ax^2 + Bxy + Cy^2 + Dx + Ey + F_1 = 0\)   ...(3)
   
   where
   
   \(D = -2Ax_c - By_c\)   ...(4)
   
   \(E = -2Cy_c - Bx_c\)   ...(5)
   
   \(F_1= A{x_c}^2 + Bx_cy_c + C{y_c}^2 + F\)   ...(6)
   
   The equations (2) and (3) above give the Equation of the Central Conic Section whose Center is Translated by an Offset \((x_c,y_c)\) from the Origin.
   
   Now, the equations (4) & (5) above can be written in form of a Matrix Equation as follows
   
   \(\begin{bmatrix}-2A & -B\\-B & -2C\end{bmatrix}\begin{bmatrix}x_c\\y_c\end{bmatrix}=\begin{bmatrix}D\\E\end{bmatrix}\)   ...(7)
   
   The Coordinates of Center of the Conic Section \((x_c,y_c)\) can be determined by solving the Matrix Equation (7) as follows
   
   \(\begin{bmatrix}x_c\\y_c\end{bmatrix}={\begin{bmatrix}-2A & -B\\-B & -2C\end{bmatrix}}^{-1}\begin{bmatrix}D\\E\end{bmatrix}\)
   
   \(\Rightarrow \begin{bmatrix}x_c\\y_c\end{bmatrix}=\frac{1}{4AC-B^2}\begin{bmatrix}-2C & B\\B & -2A\end{bmatrix}\begin{bmatrix}D\\E\end{bmatrix}\)
   
   \(\Rightarrow \begin{bmatrix}x_c\\y_c\end{bmatrix}=\begin{bmatrix}\frac{BE-2CD}{4AC-B^2}\\\frac{BD -2AE}{4AC-B^2}\end{bmatrix}\)   ...(8)