Solve r=sqrt{(r-2)^2+A^2}

$$r = \sqrt { ( r - 2 ) ^ { 2 } + A ^ { 2 } }$$

Solve for r

$$r=\frac{A^{2}+4}{4}$$

Solution Steps

Square both sides of the equation.

$$r^{2}=\left(\sqrt{\left(r-2\right)^{2}+A^{2}}\right)^{2}$$

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

$$r^{2}=\left(\sqrt{r^{2}-4r+4+A^{2}}\right)^{2}$$

Calculate \sqrt{r^{2}-4r+4+A^{2}} to the power of 2 and get r^{2}-4r+4+A^{2}.

$$r^{2}=r^{2}-4r+4+A^{2}$$

Subtract r^{2} from both sides.

$$r^{2}-r^{2}=-4r+4+A^{2}$$

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

$$0=-4r+4+A^{2}$$

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

$$-4r+4+A^{2}=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$-4r+A^{2}=-4$$

Subtract A^{2} from both sides.

$$-4r=-4-A^{2}$$

The equation is in standard form.

$$-4r=-A^{2}-4$$

Divide both sides by -4.

$$\frac{-4r}{-4}=\frac{-A^{2}-4}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$r=\frac{-A^{2}-4}{-4}$$

Divide -4-A^{2} by -4.

$$r=\frac{A^{2}}{4}+1$$

Substitute \frac{A^{2}}{4}+1 for r in the equation r=\sqrt{\left(r-2\right)^{2}+A^{2}}.

$$\frac{A^{2}}{4}+1=\sqrt{\left(\frac{A^{2}}{4}+1-2\right)^{2}+A^{2}}$$

Simplify. The value r=\frac{A^{2}}{4}+1 satisfies the equation.

$$\frac{1}{4}A^{2}+1=\frac{1}{4}\left(16+8A^{2}+A^{4}\right)^{\frac{1}{2}}$$

Equation r=\sqrt{\left(r-2\right)^{2}+A^{2}} has a unique solution.

$$r=\frac{A^{2}}{4}+1$$

Solve for A

$$A=2\sqrt{r-1}$$

$$A=-2\sqrt{r-1}\text{, }r\geq 1$$