Solve {left(-2sqrt{1/4-x^2}x+2right)}^2+z^2={left(4-z+2sqrt{1/4-x^2}x-2right)}^2

$${ \left(-2 \sqrt{ \frac{ 1 }{ 4- { x }^{ 2 } } } x+2 \right) }^{ 2 } + { z }^{ 2 } = { \left(4-z+2 \sqrt{ \frac{ 1 }{ 4- { x }^{ 2 } } } x-2 \right) }^{ 2 }$$

Solve for z

$$z=\frac{4x}{\sqrt{4-x^{2}}+x}$$

$$x\neq -\sqrt{2}\text{ and }|x|<2$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-2\sqrt{\frac{1}{4-x^{2}}}x+2\right)^{2}.

$$4\left(\sqrt{\frac{1}{4-x^{2}}}\right)^{2}x^{2}-8\sqrt{\frac{1}{4-x^{2}}}x+4+z^{2}=\left(4-z+2\sqrt{\frac{1}{4-x^{2}}}x-2\right)^{2}$$

Calculate \sqrt{\frac{1}{4-x^{2}}} to the power of 2 and get \frac{1}{4-x^{2}}.

$$4\times \frac{1}{4-x^{2}}x^{2}-8\sqrt{\frac{1}{4-x^{2}}}x+4+z^{2}=\left(4-z+2\sqrt{\frac{1}{4-x^{2}}}x-2\right)^{2}$$

Express 4\times \frac{1}{4-x^{2}} as a single fraction.

$$\frac{4}{4-x^{2}}x^{2}-8\sqrt{\frac{1}{4-x^{2}}}x+4+z^{2}=\left(4-z+2\sqrt{\frac{1}{4-x^{2}}}x-2\right)^{2}$$

Express \frac{4}{4-x^{2}}x^{2} as a single fraction.

$$\frac{4x^{2}}{4-x^{2}}-8\sqrt{\frac{1}{4-x^{2}}}x+4+z^{2}=\left(4-z+2\sqrt{\frac{1}{4-x^{2}}}x-2\right)^{2}$$

Factor 4-x^{2}.

$$\frac{4x^{2}}{\left(x-2\right)\left(-x-2\right)}-8\sqrt{\frac{1}{4-x^{2}}}x+4+z^{2}=\left(4-z+2\sqrt{\frac{1}{4-x^{2}}}x-2\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 4 times \frac{\left(x-2\right)\left(-x-2\right)}{\left(x-2\right)\left(-x-2\right)}.

$$\frac{4x^{2}}{\left(x-2\right)\left(-x-2\right)}-8\sqrt{\frac{1}{4-x^{2}}}x+\frac{4\left(x-2\right)\left(-x-2\right)}{\left(x-2\right)\left(-x-2\right)}+z^{2}=\left(4-z+2\sqrt{\frac{1}{4-x^{2}}}x-2\right)^{2}$$

Since \frac{4x^{2}}{\left(x-2\right)\left(-x-2\right)} and \frac{4\left(x-2\right)\left(-x-2\right)}{\left(x-2\right)\left(-x-2\right)} have the same denominator, add them by adding their numerators.

$$\frac{4x^{2}+4\left(x-2\right)\left(-x-2\right)}{\left(x-2\right)\left(-x-2\right)}-8\sqrt{\frac{1}{4-x^{2}}}x+z^{2}=\left(4-z+2\sqrt{\frac{1}{4-x^{2}}}x-2\right)^{2}$$

Do the multiplications in 4x^{2}+4\left(x-2\right)\left(-x-2\right).

$$\frac{4x^{2}-4x^{2}-8x+8x+16}{\left(x-2\right)\left(-x-2\right)}-8\sqrt{\frac{1}{4-x^{2}}}x+z^{2}=\left(4-z+2\sqrt{\frac{1}{4-x^{2}}}x-2\right)^{2}$$

Combine like terms in 4x^{2}-4x^{2}-8x+8x+16.

$$\frac{16}{\left(x-2\right)\left(-x-2\right)}-8\sqrt{\frac{1}{4-x^{2}}}x+z^{2}=\left(4-z+2\sqrt{\frac{1}{4-x^{2}}}x-2\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply z^{2} times \frac{\left(x-2\right)\left(-x-2\right)}{\left(x-2\right)\left(-x-2\right)}.

$$\frac{16}{\left(x-2\right)\left(-x-2\right)}-8\sqrt{\frac{1}{4-x^{2}}}x+\frac{z^{2}\left(x-2\right)\left(-x-2\right)}{\left(x-2\right)\left(-x-2\right)}=\left(4-z+2\sqrt{\frac{1}{4-x^{2}}}x-2\right)^{2}$$

Since \frac{16}{\left(x-2\right)\left(-x-2\right)} and \frac{z^{2}\left(x-2\right)\left(-x-2\right)}{\left(x-2\right)\left(-x-2\right)} have the same denominator, add them by adding their numerators.

$$\frac{16+z^{2}\left(x-2\right)\left(-x-2\right)}{\left(x-2\right)\left(-x-2\right)}-8\sqrt{\frac{1}{4-x^{2}}}x=\left(4-z+2\sqrt{\frac{1}{4-x^{2}}}x-2\right)^{2}$$

Do the multiplications in 16+z^{2}\left(x-2\right)\left(-x-2\right).

$$\frac{16-z^{2}x^{2}-2z^{2}x+2z^{2}x+4z^{2}}{\left(x-2\right)\left(-x-2\right)}-8\sqrt{\frac{1}{4-x^{2}}}x=\left(4-z+2\sqrt{\frac{1}{4-x^{2}}}x-2\right)^{2}$$

Combine like terms in 16-z^{2}x^{2}-2z^{2}x+2z^{2}x+4z^{2}.

$$\frac{4z^{2}-z^{2}x^{2}+16}{\left(x-2\right)\left(-x-2\right)}-8\sqrt{\frac{1}{4-x^{2}}}x=\left(4-z+2\sqrt{\frac{1}{4-x^{2}}}x-2\right)^{2}$$

Subtract 2 from 4 to get 2.

$$\frac{4z^{2}-z^{2}x^{2}+16}{\left(x-2\right)\left(-x-2\right)}-8\sqrt{\frac{1}{4-x^{2}}}x=\left(2-z+2\sqrt{\frac{1}{4-x^{2}}}x\right)^{2}$$

Square 2-z+2\sqrt{\frac{1}{4-x^{2}}}x.

$$\frac{4z^{2}-z^{2}x^{2}+16}{\left(x-2\right)\left(-x-2\right)}-8\sqrt{\frac{1}{4-x^{2}}}x=z^{2}-4\sqrt{\frac{1}{-x^{2}+4}}xz-4z+4\left(\sqrt{\frac{1}{-x^{2}+4}}\right)^{2}x^{2}+8\sqrt{\frac{1}{-x^{2}+4}}x+4$$

Calculate \sqrt{\frac{1}{-x^{2}+4}} to the power of 2 and get \frac{1}{-x^{2}+4}.

$$\frac{4z^{2}-z^{2}x^{2}+16}{\left(x-2\right)\left(-x-2\right)}-8\sqrt{\frac{1}{4-x^{2}}}x=z^{2}-4\sqrt{\frac{1}{-x^{2}+4}}xz-4z+4\times \frac{1}{-x^{2}+4}x^{2}+8\sqrt{\frac{1}{-x^{2}+4}}x+4$$

Express 4\times \frac{1}{-x^{2}+4} as a single fraction.

$$\frac{4z^{2}-z^{2}x^{2}+16}{\left(x-2\right)\left(-x-2\right)}-8\sqrt{\frac{1}{4-x^{2}}}x=z^{2}-4\sqrt{\frac{1}{-x^{2}+4}}xz-4z+\frac{4}{-x^{2}+4}x^{2}+8\sqrt{\frac{1}{-x^{2}+4}}x+4$$

Express \frac{4}{-x^{2}+4}x^{2} as a single fraction.

$$\frac{4z^{2}-z^{2}x^{2}+16}{\left(x-2\right)\left(-x-2\right)}-8\sqrt{\frac{1}{4-x^{2}}}x=z^{2}-4\sqrt{\frac{1}{-x^{2}+4}}xz-4z+\frac{4x^{2}}{-x^{2}+4}+8\sqrt{\frac{1}{-x^{2}+4}}x+4$$

Factor -x^{2}+4.

$$\frac{4z^{2}-z^{2}x^{2}+16}{\left(x-2\right)\left(-x-2\right)}-8\sqrt{\frac{1}{4-x^{2}}}x=z^{2}-4z+4-4\sqrt{\frac{1}{-x^{2}+4}}xz+\frac{4x^{2}}{\left(x-2\right)\left(-x-2\right)}+8\sqrt{\frac{1}{-x^{2}+4}}x$$

To add or subtract expressions, expand them to make their denominators the same. Multiply z^{2}-4z+4 times \frac{\left(x-2\right)\left(-x-2\right)}{\left(x-2\right)\left(-x-2\right)}.

$$\frac{4z^{2}-z^{2}x^{2}+16}{\left(x-2\right)\left(-x-2\right)}-8\sqrt{\frac{1}{4-x^{2}}}x=\frac{\left(z^{2}-4z+4\right)\left(x-2\right)\left(-x-2\right)}{\left(x-2\right)\left(-x-2\right)}-4\sqrt{\frac{1}{-x^{2}+4}}xz+\frac{4x^{2}}{\left(x-2\right)\left(-x-2\right)}+8\sqrt{\frac{1}{-x^{2}+4}}x$$

Since \frac{\left(z^{2}-4z+4\right)\left(x-2\right)\left(-x-2\right)}{\left(x-2\right)\left(-x-2\right)} and \frac{4x^{2}}{\left(x-2\right)\left(-x-2\right)} have the same denominator, add them by adding their numerators.

$$\frac{4z^{2}-z^{2}x^{2}+16}{\left(x-2\right)\left(-x-2\right)}-8\sqrt{\frac{1}{4-x^{2}}}x=\frac{\left(z^{2}-4z+4\right)\left(x-2\right)\left(-x-2\right)+4x^{2}}{\left(x-2\right)\left(-x-2\right)}-4\sqrt{\frac{1}{-x^{2}+4}}xz+8\sqrt{\frac{1}{-x^{2}+4}}x$$

Do the multiplications in \left(z^{2}-4z+4\right)\left(x-2\right)\left(-x-2\right)+4x^{2}.

$$\frac{4z^{2}-z^{2}x^{2}+16}{\left(x-2\right)\left(-x-2\right)}-8\sqrt{\frac{1}{4-x^{2}}}x=\frac{4z^{2}-z^{2}x^{2}-16z+4zx^{2}-4x^{2}+16+4x^{2}}{\left(x-2\right)\left(-x-2\right)}-4\sqrt{\frac{1}{-x^{2}+4}}xz+8\sqrt{\frac{1}{-x^{2}+4}}x$$

Combine like terms in 4z^{2}-z^{2}x^{2}-16z+4zx^{2}-4x^{2}+16+4x^{2}.

$$\frac{4z^{2}-z^{2}x^{2}+16}{\left(x-2\right)\left(-x-2\right)}-8\sqrt{\frac{1}{4-x^{2}}}x=\frac{16-z^{2}x^{2}+4z^{2}+4zx^{2}-16z}{\left(x-2\right)\left(-x-2\right)}-4\sqrt{\frac{1}{-x^{2}+4}}xz+8\sqrt{\frac{1}{-x^{2}+4}}x$$

Use the distributive property to multiply x-2 by -x-2 and combine like terms.

$$\frac{4z^{2}-z^{2}x^{2}+16}{-x^{2}+4}-8\sqrt{\frac{1}{4-x^{2}}}x=\frac{16-z^{2}x^{2}+4z^{2}+4zx^{2}-16z}{\left(x-2\right)\left(-x-2\right)}-4\sqrt{\frac{1}{-x^{2}+4}}xz+8\sqrt{\frac{1}{-x^{2}+4}}x$$

Use the distributive property to multiply x-2 by -x-2 and combine like terms.

$$\frac{4z^{2}-z^{2}x^{2}+16}{-x^{2}+4}-8\sqrt{\frac{1}{4-x^{2}}}x=\frac{16-z^{2}x^{2}+4z^{2}+4zx^{2}-16z}{-x^{2}+4}-4\sqrt{\frac{1}{-x^{2}+4}}xz+8\sqrt{\frac{1}{-x^{2}+4}}x$$

Subtract \frac{16-z^{2}x^{2}+4z^{2}+4zx^{2}-16z}{-x^{2}+4} from both sides.

$$\frac{4z^{2}-z^{2}x^{2}+16}{-x^{2}+4}-8\sqrt{\frac{1}{4-x^{2}}}x-\frac{16-z^{2}x^{2}+4z^{2}+4zx^{2}-16z}{-x^{2}+4}=-4\sqrt{\frac{1}{-x^{2}+4}}xz+8\sqrt{\frac{1}{-x^{2}+4}}x$$

Since \frac{4z^{2}-z^{2}x^{2}+16}{-x^{2}+4} and \frac{16-z^{2}x^{2}+4z^{2}+4zx^{2}-16z}{-x^{2}+4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4z^{2}-z^{2}x^{2}+16-\left(16-z^{2}x^{2}+4z^{2}+4zx^{2}-16z\right)}{-x^{2}+4}-8\sqrt{\frac{1}{4-x^{2}}}x=-4\sqrt{\frac{1}{-x^{2}+4}}xz+8\sqrt{\frac{1}{-x^{2}+4}}x$$

Do the multiplications in 4z^{2}-z^{2}x^{2}+16-\left(16-z^{2}x^{2}+4z^{2}+4zx^{2}-16z\right).

$$\frac{4z^{2}-z^{2}x^{2}+16-16+z^{2}x^{2}-4z^{2}-4zx^{2}+16z}{-x^{2}+4}-8\sqrt{\frac{1}{4-x^{2}}}x=-4\sqrt{\frac{1}{-x^{2}+4}}xz+8\sqrt{\frac{1}{-x^{2}+4}}x$$

Combine like terms in 4z^{2}-z^{2}x^{2}+16-16+z^{2}x^{2}-4z^{2}-4zx^{2}+16z.

$$\frac{16z-4zx^{2}}{-x^{2}+4}-8\sqrt{\frac{1}{4-x^{2}}}x=-4\sqrt{\frac{1}{-x^{2}+4}}xz+8\sqrt{\frac{1}{-x^{2}+4}}x$$

Factor the expressions that are not already factored in \frac{16z-4zx^{2}}{-x^{2}+4}.

$$\frac{4z\left(x-2\right)\left(-x-2\right)}{\left(x-2\right)\left(-x-2\right)}-8\sqrt{\frac{1}{4-x^{2}}}x=-4\sqrt{\frac{1}{-x^{2}+4}}xz+8\sqrt{\frac{1}{-x^{2}+4}}x$$

Cancel out \left(x-2\right)\left(-x-2\right) in both numerator and denominator.

$$4z-8\sqrt{\frac{1}{4-x^{2}}}x=-4\sqrt{\frac{1}{-x^{2}+4}}xz+8\sqrt{\frac{1}{-x^{2}+4}}x$$

Add 4\sqrt{\frac{1}{-x^{2}+4}}xz to both sides.

$$4z-8\sqrt{\frac{1}{4-x^{2}}}x+4\sqrt{\frac{1}{-x^{2}+4}}xz=8\sqrt{\frac{1}{-x^{2}+4}}x$$

Add 8\sqrt{\frac{1}{4-x^{2}}}x to both sides.

$$4z+4\sqrt{\frac{1}{-x^{2}+4}}xz=8\sqrt{\frac{1}{-x^{2}+4}}x+8\sqrt{\frac{1}{4-x^{2}}}x$$

Combine 8\sqrt{\frac{1}{-x^{2}+4}}x and 8\sqrt{\frac{1}{4-x^{2}}}x to get 16\sqrt{\frac{1}{-x^{2}+4}}x.

$$4z+4\sqrt{\frac{1}{-x^{2}+4}}xz=16\sqrt{\frac{1}{-x^{2}+4}}x$$

Combine all terms containing z.

$$\left(4+4\sqrt{\frac{1}{-x^{2}+4}}x\right)z=16\sqrt{\frac{1}{-x^{2}+4}}x$$

The equation is in standard form.

$$\left(4\sqrt{\frac{1}{4-x^{2}}}x+4\right)z=16\sqrt{\frac{1}{4-x^{2}}}x$$

Divide both sides by 4+4\sqrt{\left(-x^{2}+4\right)^{-1}}x.

$$\frac{\left(4\sqrt{\frac{1}{4-x^{2}}}x+4\right)z}{4\sqrt{\frac{1}{4-x^{2}}}x+4}=\frac{16x}{\sqrt{4-x^{2}}\left(4\sqrt{\frac{1}{4-x^{2}}}x+4\right)}$$

Dividing by 4+4\sqrt{\left(-x^{2}+4\right)^{-1}}x undoes the multiplication by 4+4\sqrt{\left(-x^{2}+4\right)^{-1}}x.

$$z=\frac{16x}{\sqrt{4-x^{2}}\left(4\sqrt{\frac{1}{4-x^{2}}}x+4\right)}$$

Divide \frac{16x}{\sqrt{4-x^{2}}} by 4+4\sqrt{\left(-x^{2}+4\right)^{-1}}x.

$$z=\frac{4x}{\sqrt{4-x^{2}}+x}$$

Solve for x

$$\left\{\begin{matrix}x=-\sqrt{\frac{2}{z^{2}-4z+8}}z\text{, }&z=0\text{ or }z>4\\x=\sqrt{\frac{2}{z^{2}-4z+8}}z\text{, }&z<4\end{matrix}\right.$$