Solve x^2+y^2+2=2az

$$x ^ { 2 } + y ^ { 2 } + 2 = 2 a z$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{x^{2}+y^{2}+2}{2z}\text{, }&z\neq 0\\a\in \mathrm{C}\text{, }&\left(x=i\sqrt{y^{2}+2}\text{ or }x=-i\sqrt{y^{2}+2}\right)\text{ and }z=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$2az=x^{2}+y^{2}+2$$

The equation is in standard form.

$$2za=x^{2}+y^{2}+2$$

Divide both sides by 2z.

$$\frac{2za}{2z}=\frac{x^{2}+y^{2}+2}{2z}$$

Dividing by 2z undoes the multiplication by 2z.

$$a=\frac{x^{2}+y^{2}+2}{2z}$$

Solve for a

$$a=\frac{x^{2}+y^{2}+2}{2z}$$

$$z\neq 0$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$2az=x^{2}+y^{2}+2$$

The equation is in standard form.

$$2za=x^{2}+y^{2}+2$$

Divide both sides by 2z.

$$\frac{2za}{2z}=\frac{x^{2}+y^{2}+2}{2z}$$

Dividing by 2z undoes the multiplication by 2z.

$$a=\frac{x^{2}+y^{2}+2}{2z}$$

Solve for x (complex solution)

$$x=-i\sqrt{y^{2}-2az+2}$$

$$x=i\sqrt{y^{2}-2az+2}$$