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

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

Solve for x

$$x=\frac{y-3}{2}$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

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

$$x^{2}+y^{2}+4y+4=\left(\sqrt{x^{2}+8x+16+y^{2}}\right)^{2}$$

Calculate \sqrt{x^{2}+8x+16+y^{2}} to the power of 2 and get x^{2}+8x+16+y^{2}.

$$x^{2}+y^{2}+4y+4=x^{2}+8x+16+y^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+y^{2}+4y+4-x^{2}=8x+16+y^{2}$$

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

$$y^{2}+4y+4=8x+16+y^{2}$$

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

$$8x+16+y^{2}=y^{2}+4y+4$$

Subtract 16 from both sides.

$$8x+y^{2}=y^{2}+4y+4-16$$

Subtract 16 from 4 to get -12.

$$8x+y^{2}=y^{2}+4y-12$$

Subtract y^{2} from both sides.

$$8x=y^{2}+4y-12-y^{2}$$

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

$$8x=4y-12$$

Divide both sides by 8.

$$\frac{8x}{8}=\frac{4y-12}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x=\frac{4y-12}{8}$$

Divide -12+4y by 8.

$$x=\frac{y-3}{2}$$

Substitute \frac{y-3}{2} for x in the equation \sqrt{x^{2}+\left(y+2\right)^{2}}=\sqrt{\left(x+4\right)^{2}+y^{2}}.

$$\sqrt{\left(\frac{y-3}{2}\right)^{2}+\left(y+2\right)^{2}}=\sqrt{\left(\frac{y-3}{2}+4\right)^{2}+y^{2}}$$

Simplify. The value x=\frac{y-3}{2} satisfies the equation.

$$\frac{1}{2}\left(25+10y+5y^{2}\right)^{\frac{1}{2}}=\frac{1}{2}\left(\left(y+5\right)^{2}+4y^{2}\right)^{\frac{1}{2}}$$

Equation \sqrt{\left(y+2\right)^{2}+x^{2}}=\sqrt{\left(x+4\right)^{2}+y^{2}} has a unique solution.

$$x=\frac{y-3}{2}$$

Solve for y

$$y=2x+3$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

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

$$x^{2}+y^{2}+4y+4=\left(\sqrt{x^{2}+8x+16+y^{2}}\right)^{2}$$

Calculate \sqrt{x^{2}+8x+16+y^{2}} to the power of 2 and get x^{2}+8x+16+y^{2}.

$$x^{2}+y^{2}+4y+4=x^{2}+8x+16+y^{2}$$

Subtract y^{2} from both sides.

$$x^{2}+y^{2}+4y+4-y^{2}=x^{2}+8x+16$$

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

$$x^{2}+4y+4=x^{2}+8x+16$$

Subtract x^{2} from both sides.

$$4y+4=x^{2}+8x+16-x^{2}$$

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

$$4y+4=8x+16$$

Subtract 4 from both sides.

$$4y=8x+16-4$$

Subtract 4 from 16 to get 12.

$$4y=8x+12$$

Divide both sides by 4.

$$\frac{4y}{4}=\frac{8x+12}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$y=\frac{8x+12}{4}$$

Divide 8x+12 by 4.

$$y=2x+3$$

Substitute 2x+3 for y in the equation \sqrt{x^{2}+\left(y+2\right)^{2}}=\sqrt{\left(x+4\right)^{2}+y^{2}}.

$$\sqrt{x^{2}+\left(2x+3+2\right)^{2}}=\sqrt{\left(x+4\right)^{2}+\left(2x+3\right)^{2}}$$

Simplify. The value y=2x+3 satisfies the equation.

$$\left(x^{2}+\left(2x+5\right)^{2}\right)^{\frac{1}{2}}=\left(25+20x+5x^{2}\right)^{\frac{1}{2}}$$

Equation \sqrt{\left(y+2\right)^{2}+x^{2}}=\sqrt{\left(x+4\right)^{2}+y^{2}} has a unique solution.

$$y=2x+3$$