Solve (y-x)^2+5^2=y^2

$$( y - x ) ^ { 2 } + 5 ^ { 2 } = y ^ { 2 }$$

Solve for y

$$y=\frac{x}{2}+\frac{25}{2x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

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

$$y^{2}-2yx+x^{2}+5^{2}=y^{2}$$

Calculate 5 to the power of 2 and get 25.

$$y^{2}-2yx+x^{2}+25=y^{2}$$

Subtract y^{2} from both sides.

$$y^{2}-2yx+x^{2}+25-y^{2}=0$$

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

$$-2yx+x^{2}+25=0$$

Subtract x^{2} from both sides. Anything subtracted from zero gives its negation.

$$-2yx+25=-x^{2}$$

Subtract 25 from both sides.

$$-2yx=-x^{2}-25$$

The equation is in standard form.

$$\left(-2x\right)y=-x^{2}-25$$

Divide both sides by -2x.

$$\frac{\left(-2x\right)y}{-2x}=\frac{-x^{2}-25}{-2x}$$

Dividing by -2x undoes the multiplication by -2x.

$$y=\frac{-x^{2}-25}{-2x}$$

Divide -x^{2}-25 by -2x.

$$y=\frac{x}{2}+\frac{25}{2x}$$

Solve for x (complex solution)

$$x=\sqrt{y^{2}-25}+y$$

$$x=-\sqrt{y^{2}-25}+y$$

Solve for x

$$x=\sqrt{y^{2}-25}+y$$

$$x=-\sqrt{y^{2}-25}+y\text{, }|y|\geq 5$$