Solve (1/2-h)^2+(2-k)^2-(1+h)^2-(1+K)^2=0

$$( \frac { 1 } { 2 } - h ) ^ { 2 } + ( 2 - k ) ^ { 2 } - ( 1 + h ) ^ { 2 } - ( 1 + K ) ^ { 2 } = 0$$

Solve for h

$$h=\frac{k^{2}}{3}-\frac{K^{2}}{3}-\frac{2K}{3}-\frac{4k}{3}+\frac{3}{4}$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{2}-h\right)^{2}.

$$\frac{1}{4}-h+h^{2}+\left(2-k\right)^{2}-\left(1+h\right)^{2}-\left(1+K\right)^{2}=0$$

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

$$\frac{1}{4}-h+h^{2}+4-4k+k^{2}-\left(1+h\right)^{2}-\left(1+K\right)^{2}=0$$

Add \frac{1}{4} and 4 to get \frac{17}{4}.

$$\frac{17}{4}-h+h^{2}-4k+k^{2}-\left(1+h\right)^{2}-\left(1+K\right)^{2}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+h\right)^{2}.

$$\frac{17}{4}-h+h^{2}-4k+k^{2}-\left(1+2h+h^{2}\right)-\left(1+K\right)^{2}=0$$

To find the opposite of 1+2h+h^{2}, find the opposite of each term.

$$\frac{17}{4}-h+h^{2}-4k+k^{2}-1-2h-h^{2}-\left(1+K\right)^{2}=0$$

Subtract 1 from \frac{17}{4} to get \frac{13}{4}.

$$\frac{13}{4}-h+h^{2}-4k+k^{2}-2h-h^{2}-\left(1+K\right)^{2}=0$$

Combine -h and -2h to get -3h.

$$\frac{13}{4}-3h+h^{2}-4k+k^{2}-h^{2}-\left(1+K\right)^{2}=0$$

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

$$\frac{13}{4}-3h-4k+k^{2}-\left(1+K\right)^{2}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+K\right)^{2}.

$$\frac{13}{4}-3h-4k+k^{2}-\left(1+2K+K^{2}\right)=0$$

To find the opposite of 1+2K+K^{2}, find the opposite of each term.

$$\frac{13}{4}-3h-4k+k^{2}-1-2K-K^{2}=0$$

Subtract 1 from \frac{13}{4} to get \frac{9}{4}.

$$\frac{9}{4}-3h-4k+k^{2}-2K-K^{2}=0$$

Subtract \frac{9}{4} from both sides. Anything subtracted from zero gives its negation.

$$-3h-4k+k^{2}-2K-K^{2}=-\frac{9}{4}$$

Add 4k to both sides.

$$-3h+k^{2}-2K-K^{2}=-\frac{9}{4}+4k$$

Subtract k^{2} from both sides.

$$-3h-2K-K^{2}=-\frac{9}{4}+4k-k^{2}$$

Add 2K to both sides.

$$-3h-K^{2}=-\frac{9}{4}+4k-k^{2}+2K$$

Add K^{2} to both sides.

$$-3h=-\frac{9}{4}+4k-k^{2}+2K+K^{2}$$

The equation is in standard form.

$$-3h=K^{2}+2K-k^{2}+4k-\frac{9}{4}$$

Divide both sides by -3.

$$\frac{-3h}{-3}=\frac{K^{2}+2K-k^{2}+4k-\frac{9}{4}}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$h=\frac{K^{2}+2K-k^{2}+4k-\frac{9}{4}}{-3}$$

Divide -\frac{9}{4}+4k-k^{2}+2K+K^{2} by -3.

$$h=\frac{k^{2}}{3}-\frac{K^{2}}{3}-\frac{2K}{3}-\frac{4k}{3}+\frac{3}{4}$$

Solve for K (complex solution)

$$K=-\frac{\sqrt{13-16k+4k^{2}-12h}}{2}-1$$

$$K=\frac{\sqrt{13-16k+4k^{2}-12h}}{2}-1$$

Solve for K

$$K=-\frac{\sqrt{13-16k+4k^{2}-12h}}{2}-1$$

$$K=\frac{\sqrt{13-16k+4k^{2}-12h}}{2}-1\text{, }h\leq \frac{k^{2}}{3}-\frac{4k}{3}+\frac{13}{12}$$