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

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

Solve for x

$$x=\frac{y-1}{3}$$

Solution Steps

Square both sides of the equation.

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

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

$$\left(\sqrt{x^{2}+6x+9+\left(y-2\right)^{2}}\right)^{2}=\left(\sqrt{\left(x-3\right)^{2}+y^{2}}\right)^{2}$$

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

$$\left(\sqrt{x^{2}+6x+9+y^{2}-4y+4}\right)^{2}=\left(\sqrt{\left(x-3\right)^{2}+y^{2}}\right)^{2}$$

Add 9 and 4 to get 13.

$$\left(\sqrt{x^{2}+6x+13+y^{2}-4y}\right)^{2}=\left(\sqrt{\left(x-3\right)^{2}+y^{2}}\right)^{2}$$

Calculate \sqrt{x^{2}+6x+13+y^{2}-4y} to the power of 2 and get x^{2}+6x+13+y^{2}-4y.

$$x^{2}+6x+13+y^{2}-4y=\left(\sqrt{\left(x-3\right)^{2}+y^{2}}\right)^{2}$$

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

$$x^{2}+6x+13+y^{2}-4y=\left(\sqrt{x^{2}-6x+9+y^{2}}\right)^{2}$$

Calculate \sqrt{x^{2}-6x+9+y^{2}} to the power of 2 and get x^{2}-6x+9+y^{2}.

$$x^{2}+6x+13+y^{2}-4y=x^{2}-6x+9+y^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+6x+13+y^{2}-4y-x^{2}=-6x+9+y^{2}$$

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

$$6x+13+y^{2}-4y=-6x+9+y^{2}$$

Add 6x to both sides.

$$6x+13+y^{2}-4y+6x=9+y^{2}$$

Combine 6x and 6x to get 12x.

$$12x+13+y^{2}-4y=9+y^{2}$$

Subtract 13 from both sides.

$$12x+y^{2}-4y=9+y^{2}-13$$

Subtract 13 from 9 to get -4.

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

Subtract y^{2} from both sides.

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

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

$$12x-4y=-4$$

Add 4y to both sides.

$$12x=-4+4y$$

The equation is in standard form.

$$12x=4y-4$$

Divide both sides by 12.

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

Dividing by 12 undoes the multiplication by 12.

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

Divide -4+4y by 12.

$$x=\frac{y-1}{3}$$

Substitute \frac{y-1}{3} for x in the equation \sqrt{\left(x+3\right)^{2}+\left(y-2\right)^{2}}=\sqrt{\left(x-3\right)^{2}+y^{2}}.

$$\sqrt{\left(\frac{y-1}{3}+3\right)^{2}+\left(y-2\right)^{2}}=\sqrt{\left(\frac{y-1}{3}-3\right)^{2}+y^{2}}$$

Simplify. The value x=\frac{y-1}{3} satisfies the equation.

$$\frac{1}{3}\left(100-20y+10y^{2}\right)^{\frac{1}{2}}=\frac{1}{3}\left(\left(y-10\right)^{2}+9y^{2}\right)^{\frac{1}{2}}$$

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

$$x=\frac{y-1}{3}$$

Solve for y

$$y=3x+1$$

Solution Steps

Square both sides of the equation.

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

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

$$\left(\sqrt{x^{2}+6x+9+\left(y-2\right)^{2}}\right)^{2}=\left(\sqrt{\left(x-3\right)^{2}+y^{2}}\right)^{2}$$

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

$$\left(\sqrt{x^{2}+6x+9+y^{2}-4y+4}\right)^{2}=\left(\sqrt{\left(x-3\right)^{2}+y^{2}}\right)^{2}$$

Add 9 and 4 to get 13.

$$\left(\sqrt{x^{2}+6x+13+y^{2}-4y}\right)^{2}=\left(\sqrt{\left(x-3\right)^{2}+y^{2}}\right)^{2}$$

Calculate \sqrt{x^{2}+6x+13+y^{2}-4y} to the power of 2 and get x^{2}+6x+13+y^{2}-4y.

$$x^{2}+6x+13+y^{2}-4y=\left(\sqrt{\left(x-3\right)^{2}+y^{2}}\right)^{2}$$

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

$$x^{2}+6x+13+y^{2}-4y=\left(\sqrt{x^{2}-6x+9+y^{2}}\right)^{2}$$

Calculate \sqrt{x^{2}-6x+9+y^{2}} to the power of 2 and get x^{2}-6x+9+y^{2}.

$$x^{2}+6x+13+y^{2}-4y=x^{2}-6x+9+y^{2}$$

Subtract y^{2} from both sides.

$$x^{2}+6x+13+y^{2}-4y-y^{2}=x^{2}-6x+9$$

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

$$x^{2}+6x+13-4y=x^{2}-6x+9$$

Subtract x^{2} from both sides.

$$6x+13-4y=x^{2}-6x+9-x^{2}$$

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

$$6x+13-4y=-6x+9$$

Subtract 6x from both sides.

$$13-4y=-6x+9-6x$$

Combine -6x and -6x to get -12x.

$$13-4y=-12x+9$$

Subtract 13 from both sides.

$$-4y=-12x+9-13$$

Subtract 13 from 9 to get -4.

$$-4y=-12x-4$$

Divide both sides by -4.

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

Dividing by -4 undoes the multiplication by -4.

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

Divide -12x-4 by -4.

$$y=3x+1$$

Substitute 3x+1 for y in the equation \sqrt{\left(x+3\right)^{2}+\left(y-2\right)^{2}}=\sqrt{\left(x-3\right)^{2}+y^{2}}.

$$\sqrt{\left(x+3\right)^{2}+\left(3x+1-2\right)^{2}}=\sqrt{\left(x-3\right)^{2}+\left(3x+1\right)^{2}}$$

Simplify. The value y=3x+1 satisfies the equation.

$$\left(10+10x^{2}\right)^{\frac{1}{2}}=\left(10+10x^{2}\right)^{\frac{1}{2}}$$

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

$$y=3x+1$$