Solve [x-(p-b)]^2+(y-p)^2=2p^2-4p_{m}

$$[ x - ( p - b ) ] ^ { 2 } + ( y - p ) ^ { 2 } = 2 p ^ { 2 } - 4 p _ { m }$$

Solve for p (complex solution)

$$\left\{\begin{matrix}p=\frac{x^{2}+2bx+y^{2}+b^{2}+4p_{m}}{2\left(x+y+b\right)}\text{, }&x\neq -y-b\\p\in \mathrm{C}\text{, }&p_{m}=-\frac{y^{2}}{2}\text{ and }x=-\left(y+b\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

To find the opposite of p-b, find the opposite of each term.

$$\left(x-p+b\right)^{2}+\left(y-p\right)^{2}=2p^{2}-4p_{m}$$

Square x-p+b.

$$x^{2}+2bx-2px-2bp+b^{2}+p^{2}+\left(y-p\right)^{2}=2p^{2}-4p_{m}$$

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

$$x^{2}+2bx-2px-2bp+b^{2}+p^{2}+y^{2}-2yp+p^{2}=2p^{2}-4p_{m}$$

Combine p^{2} and p^{2} to get 2p^{2}.

$$x^{2}+2bx-2px-2bp+b^{2}+2p^{2}+y^{2}-2yp=2p^{2}-4p_{m}$$

Subtract 2p^{2} from both sides.

$$x^{2}+2bx-2px-2bp+b^{2}+2p^{2}+y^{2}-2yp-2p^{2}=-4p_{m}$$

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

$$x^{2}+2bx-2px-2bp+b^{2}+y^{2}-2yp=-4p_{m}$$

Subtract x^{2} from both sides.

$$2bx-2px-2bp+b^{2}+y^{2}-2yp=-4p_{m}-x^{2}$$

Subtract 2bx from both sides.

$$-2px-2bp+b^{2}+y^{2}-2yp=-4p_{m}-x^{2}-2bx$$

Subtract b^{2} from both sides.

$$-2px-2bp+y^{2}-2yp=-4p_{m}-x^{2}-2bx-b^{2}$$

Subtract y^{2} from both sides.

$$-2px-2bp-2yp=-4p_{m}-x^{2}-2bx-b^{2}-y^{2}$$

Combine all terms containing p.

$$\left(-2x-2b-2y\right)p=-4p_{m}-x^{2}-2bx-b^{2}-y^{2}$$

The equation is in standard form.

$$\left(-2x-2y-2b\right)p=-x^{2}-2bx-y^{2}-b^{2}-4p_{m}$$

Divide both sides by -2x-2b-2y.

$$\frac{\left(-2x-2y-2b\right)p}{-2x-2y-2b}=\frac{-\left(x+b\right)^{2}-y^{2}-4p_{m}}{-2x-2y-2b}$$

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

$$p=\frac{-\left(x+b\right)^{2}-y^{2}-4p_{m}}{-2x-2y-2b}$$

Divide -4p_{m}-y^{2}-\left(x+b\right)^{2} by -2x-2b-2y.

$$p=\frac{x^{2}+2bx+y^{2}+b^{2}+4p_{m}}{2\left(x+y+b\right)}$$

Solve for p

$$\left\{\begin{matrix}p=\frac{x^{2}+2bx+y^{2}+b^{2}+4p_{m}}{2\left(x+y+b\right)}\text{, }&x\neq -y-b\\p\in \mathrm{R}\text{, }&p_{m}=-\frac{y^{2}}{2}\text{ and }x=-\left(y+b\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

To find the opposite of p-b, find the opposite of each term.

$$\left(x-p+b\right)^{2}+\left(y-p\right)^{2}=2p^{2}-4p_{m}$$

Square x-p+b.

$$x^{2}+2bx-2px-2bp+b^{2}+p^{2}+\left(y-p\right)^{2}=2p^{2}-4p_{m}$$

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

$$x^{2}+2bx-2px-2bp+b^{2}+p^{2}+y^{2}-2yp+p^{2}=2p^{2}-4p_{m}$$

Combine p^{2} and p^{2} to get 2p^{2}.

$$x^{2}+2bx-2px-2bp+b^{2}+2p^{2}+y^{2}-2yp=2p^{2}-4p_{m}$$

Subtract 2p^{2} from both sides.

$$x^{2}+2bx-2px-2bp+b^{2}+2p^{2}+y^{2}-2yp-2p^{2}=-4p_{m}$$

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

$$x^{2}+2bx-2px-2bp+b^{2}+y^{2}-2yp=-4p_{m}$$

Subtract x^{2} from both sides.

$$2bx-2px-2bp+b^{2}+y^{2}-2yp=-4p_{m}-x^{2}$$

Subtract 2bx from both sides.

$$-2px-2bp+b^{2}+y^{2}-2yp=-4p_{m}-x^{2}-2bx$$

Subtract b^{2} from both sides.

$$-2px-2bp+y^{2}-2yp=-4p_{m}-x^{2}-2bx-b^{2}$$

Subtract y^{2} from both sides.

$$-2px-2bp-2yp=-4p_{m}-x^{2}-2bx-b^{2}-y^{2}$$

Combine all terms containing p.

$$\left(-2x-2b-2y\right)p=-4p_{m}-x^{2}-2bx-b^{2}-y^{2}$$

The equation is in standard form.

$$\left(-2x-2y-2b\right)p=-x^{2}-2bx-y^{2}-b^{2}-4p_{m}$$

Divide both sides by -2x-2b-2y.

$$\frac{\left(-2x-2y-2b\right)p}{-2x-2y-2b}=\frac{-\left(x+b\right)^{2}-y^{2}-4p_{m}}{-2x-2y-2b}$$

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

$$p=\frac{-\left(x+b\right)^{2}-y^{2}-4p_{m}}{-2x-2y-2b}$$

Divide -4p_{m}-y^{2}-\left(x+b\right)^{2} by -2x-2b-2y.

$$p=\frac{x^{2}+2bx+y^{2}+b^{2}+4p_{m}}{2\left(x+y+b\right)}$$

Solve for b (complex solution)

$$b=-i\sqrt{y^{2}-2py-p^{2}+4p_{m}}+p-x$$

$$b=i\sqrt{y^{2}-2py-p^{2}+4p_{m}}+p-x$$

Solve for b

$$b=-\left(\sqrt{-y^{2}+2py+p^{2}-4p_{m}}+x-p\right)$$

$$b=\sqrt{-y^{2}+2py+p^{2}-4p_{m}}+p-x\text{, }p_{m}\leq \frac{py}{2}+\frac{p^{2}}{4}-\frac{y^{2}}{4}$$