Solve sqrt{x^2+y^2}=sqrt{(x-5)^2+(y+3)^2}

$$\sqrt { x ^ { 2 } + y ^ { 2 } } = \sqrt { ( x - 5 ) ^ { 2 } + ( y + 3 ) ^ { 2 } }$$

Solve for x

$$x=\frac{3y+17}{5}$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x^{2}+y^{2}}\right)^{2}=\left(\sqrt{\left(x-5\right)^{2}+\left(y+3\right)^{2}}\right)^{2}$$

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

$$x^{2}+y^{2}=\left(\sqrt{\left(x-5\right)^{2}+\left(y+3\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}+y^{2}=\left(\sqrt{x^{2}-10x+25+\left(y+3\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+3\right)^{2}.

$$x^{2}+y^{2}=\left(\sqrt{x^{2}-10x+25+y^{2}+6y+9}\right)^{2}$$

Add 25 and 9 to get 34.

$$x^{2}+y^{2}=\left(\sqrt{x^{2}-10x+34+y^{2}+6y}\right)^{2}$$

Calculate \sqrt{x^{2}-10x+34+y^{2}+6y} to the power of 2 and get x^{2}-10x+34+y^{2}+6y.

$$x^{2}+y^{2}=x^{2}-10x+34+y^{2}+6y$$

Subtract x^{2} from both sides.

$$x^{2}+y^{2}-x^{2}=-10x+34+y^{2}+6y$$

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

$$y^{2}=-10x+34+y^{2}+6y$$

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

$$-10x+34+y^{2}+6y=y^{2}$$

Subtract 34 from both sides.

$$-10x+y^{2}+6y=y^{2}-34$$

Subtract y^{2} from both sides.

$$-10x+6y=y^{2}-34-y^{2}$$

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

$$-10x+6y=-34$$

Subtract 6y from both sides.

$$-10x=-34-6y$$

The equation is in standard form.

$$-10x=-6y-34$$

Divide both sides by -10.

$$\frac{-10x}{-10}=\frac{-6y-34}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x=\frac{-6y-34}{-10}$$

Divide -34-6y by -10.

$$x=\frac{3y+17}{5}$$

Substitute \frac{3y+17}{5} for x in the equation \sqrt{x^{2}+y^{2}}=\sqrt{\left(x-5\right)^{2}+\left(y+3\right)^{2}}.

$$\sqrt{\left(\frac{3y+17}{5}\right)^{2}+y^{2}}=\sqrt{\left(\frac{3y+17}{5}-5\right)^{2}+\left(y+3\right)^{2}}$$

Simplify. The value x=\frac{3y+17}{5} satisfies the equation.

$$\frac{1}{5}\left(34y^{2}+102y+289\right)^{\frac{1}{2}}=\frac{1}{5}\left(289+102y+34y^{2}\right)^{\frac{1}{2}}$$

Equation \sqrt{x^{2}+y^{2}}=\sqrt{\left(y+3\right)^{2}+\left(x-5\right)^{2}} has a unique solution.

$$x=\frac{3y+17}{5}$$

Solve for y

$$y=\frac{5x-17}{3}$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x^{2}+y^{2}}\right)^{2}=\left(\sqrt{\left(x-5\right)^{2}+\left(y+3\right)^{2}}\right)^{2}$$

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

$$x^{2}+y^{2}=\left(\sqrt{\left(x-5\right)^{2}+\left(y+3\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}+y^{2}=\left(\sqrt{x^{2}-10x+25+\left(y+3\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+3\right)^{2}.

$$x^{2}+y^{2}=\left(\sqrt{x^{2}-10x+25+y^{2}+6y+9}\right)^{2}$$

Add 25 and 9 to get 34.

$$x^{2}+y^{2}=\left(\sqrt{x^{2}-10x+34+y^{2}+6y}\right)^{2}$$

Calculate \sqrt{x^{2}-10x+34+y^{2}+6y} to the power of 2 and get x^{2}-10x+34+y^{2}+6y.

$$x^{2}+y^{2}=x^{2}-10x+34+y^{2}+6y$$

Subtract y^{2} from both sides.

$$x^{2}+y^{2}-y^{2}=x^{2}-10x+34+6y$$

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

$$x^{2}=x^{2}-10x+34+6y$$

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

$$x^{2}-10x+34+6y=x^{2}$$

Subtract x^{2} from both sides.

$$-10x+34+6y=x^{2}-x^{2}$$

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

$$-10x+34+6y=0$$

Add 10x to both sides. Anything plus zero gives itself.

$$34+6y=10x$$

Subtract 34 from both sides.

$$6y=10x-34$$

Divide both sides by 6.

$$\frac{6y}{6}=\frac{10x-34}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$y=\frac{10x-34}{6}$$

Divide 10x-34 by 6.

$$y=\frac{5x-17}{3}$$

Substitute \frac{5x-17}{3} for y in the equation \sqrt{x^{2}+y^{2}}=\sqrt{\left(x-5\right)^{2}+\left(y+3\right)^{2}}.

$$\sqrt{x^{2}+\left(\frac{5x-17}{3}\right)^{2}}=\sqrt{\left(x-5\right)^{2}+\left(\frac{5x-17}{3}+3\right)^{2}}$$

Simplify. The value y=\frac{5x-17}{3} satisfies the equation.

$$\frac{1}{3}\left(34x^{2}-170x+289\right)^{\frac{1}{2}}=\frac{1}{3}\left(289-170x+34x^{2}\right)^{\frac{1}{2}}$$

Equation \sqrt{x^{2}+y^{2}}=\sqrt{\left(y+3\right)^{2}+\left(x-5\right)^{2}} has a unique solution.

$$y=\frac{5x-17}{3}$$