Solve left(x+10right)^2+y=left(x-6right)^2+{left(y-4sqrt{2}right)}^2

$${ \left(x+10 \right) }^{ 2 } +y= { \left(x-6 \right) }^{ 2 } + { \left(y-4 \sqrt{ 2 } \right) }^{ 2 }$$

Solve for x

$$x=\frac{\left(y-4\sqrt{2}\right)^{2}-y-64}{32}$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+10\right)^{2}.

$$x^{2}+20x+100+y=\left(x-6\right)^{2}+\left(y-4\sqrt{2}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$x^{2}+20x+100+y=x^{2}-12x+36+\left(y-4\sqrt{2}\right)^{2}$$

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

$$x^{2}+20x+100+y=x^{2}-12x+36+y^{2}-8y\sqrt{2}+16\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$x^{2}+20x+100+y=x^{2}-12x+36+y^{2}-8y\sqrt{2}+16\times 2$$

Multiply 16 and 2 to get 32.

$$x^{2}+20x+100+y=x^{2}-12x+36+y^{2}-8y\sqrt{2}+32$$

Add 36 and 32 to get 68.

$$x^{2}+20x+100+y=x^{2}-12x+68+y^{2}-8y\sqrt{2}$$

Subtract x^{2} from both sides.

$$x^{2}+20x+100+y-x^{2}=-12x+68+y^{2}-8y\sqrt{2}$$

Combine x^{2} and -x^{2} to get 0.

$$20x+100+y=-12x+68+y^{2}-8y\sqrt{2}$$

Add 12x to both sides.

$$20x+100+y+12x=68+y^{2}-8y\sqrt{2}$$

Combine 20x and 12x to get 32x.

$$32x+100+y=68+y^{2}-8y\sqrt{2}$$

Subtract 100 from both sides.

$$32x+y=68+y^{2}-8y\sqrt{2}-100$$

Subtract 100 from 68 to get -32.

$$32x+y=-32+y^{2}-8y\sqrt{2}$$

Subtract y from both sides.

$$32x=-32+y^{2}-8y\sqrt{2}-y$$

The equation is in standard form.

$$32x=y^{2}-8\sqrt{2}y-y-32$$

Divide both sides by 32.

$$\frac{32x}{32}=\frac{y^{2}-8\sqrt{2}y-y-32}{32}$$

Dividing by 32 undoes the multiplication by 32.

$$x=\frac{y^{2}-8\sqrt{2}y-y-32}{32}$$

Divide -32+y^{2}-8y\sqrt{2}-y by 32.

$$x=\frac{y^{2}}{32}-\frac{\sqrt{2}y}{4}-\frac{y}{32}-1$$

Solve for y (complex solution)

$$y=-\frac{\sqrt{128x+16\sqrt{2}+257}}{2}+4\sqrt{2}+\frac{1}{2}$$

$$y=\frac{\sqrt{128x+16\sqrt{2}+257}}{2}+4\sqrt{2}+\frac{1}{2}$$

Solve for y

$$y=-\frac{\sqrt{128x+16\sqrt{2}+257}}{2}+4\sqrt{2}+\frac{1}{2}$$

$$y=\frac{\sqrt{128x+16\sqrt{2}+257}}{2}+4\sqrt{2}+\frac{1}{2}\text{, }x\geq -\frac{\sqrt{2}}{8}-\frac{257}{128}$$