Solve for $$ x $$ and $$ y $$ simultaneously : $$ (x-2)^{2}+(y-3)^{2}=4 \quad $$ and $$ \quad x+y=4 $$

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Solve the system of equations $$ (x-2)^{2}+(y-3)^{2}=4;x+y=4 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}\left(x-2ight)^{2}+\left(y-3ight)^{2}=4\x+y=4\end{array}ight.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}\left(x-2ight)^{2}+\left(y-3ight)^{2}=4\x=4-y\end{array}ight.$$
- step2: Substitute the value of $$x:$$
  $$\left(4-y-2ight)^{2}+\left(y-3ight)^{2}=4$$
- step3: Simplify:
  $$13-10y+2y^{2}=4$$
- step4: Move the expression to the left side:
  $$13-10y+2y^{2}-4=0$$
- step5: Subtract the numbers:
  $$9-10y+2y^{2}=0$$
- step6: Rewrite in standard form:
  $$2y^{2}-10y+9=0$$
- step7: Solve using the quadratic formula:
  $$y=\frac{10\pm \sqrt{\left(-10ight)^{2}-4	imes 2	imes 9}}{2	imes 2}$$
- step8: Simplify the expression:
  $$y=\frac{10\pm \sqrt{\left(-10ight)^{2}-4	imes 2	imes 9}}{4}$$
- step9: Simplify the expression:
  $$y=\frac{10\pm \sqrt{28}}{4}$$
- step10: Simplify the expression:
  $$y=\frac{10\pm 2\sqrt{7}}{4}$$
- step11: Separate into possible cases:
  $$\begin{align}&y=\frac{10+2\sqrt{7}}{4}\&y=\frac{10-2\sqrt{7}}{4}\end{align}$$
- step12: Simplify the expression:
  $$\begin{align}&y=\frac{5+\sqrt{7}}{2}\&y=\frac{10-2\sqrt{7}}{4}\end{align}$$
- step13: Simplify the expression:
  $$\begin{align}&y=\frac{5+\sqrt{7}}{2}\&y=\frac{5-\sqrt{7}}{2}\end{align}$$
- step14: Evaluate the logic:
  $$y=\frac{5+\sqrt{7}}{2}\cup y=\frac{5-\sqrt{7}}{2}$$
- step15: Rearrange the terms:
  $$\left\{ \begin{array}{l}x=4-y\y=\frac{5+\sqrt{7}}{2}\end{array}ight.\cup \left\{ \begin{array}{l}x=4-y\y=\frac{5-\sqrt{7}}{2}\end{array}ight.$$
- step16: Calculate:
  $$\left\{ \begin{array}{l}x=\frac{3-\sqrt{7}}{2}\y=\frac{5+\sqrt{7}}{2}\end{array}ight.\cup \left\{ \begin{array}{l}x=\frac{3+\sqrt{7}}{2}\y=\frac{5-\sqrt{7}}{2}\end{array}ight.$$
- step17: Calculate:
  $$\left\{ \begin{array}{l}x=\frac{3+\sqrt{7}}{2}\y=\frac{5-\sqrt{7}}{2}\end{array}ight.\cup \left\{ \begin{array}{l}x=\frac{3-\sqrt{7}}{2}\y=\frac{5+\sqrt{7}}{2}\end{array}ight.$$
- step18: Check the solution:
  $$\left\{ \begin{array}{l}x=\frac{3+\sqrt{7}}{2}\y=\frac{5-\sqrt{7}}{2}\end{array}ight.\cup \left\{ \begin{array}{l}x=\frac{3-\sqrt{7}}{2}\y=\frac{5+\sqrt{7}}{2}\end{array}ight.$$
- step19: Rewrite:
  $$\left(x,yight) = \left(\frac{3+\sqrt{7}}{2},\frac{5-\sqrt{7}}{2}ight)\cup \left(x,yight) = \left(\frac{3-\sqrt{7}}{2},\frac{5+\sqrt{7}}{2}ight)$$
The solutions to the system of equations $$ (x-2)^{2}+(y-3)^{2}=4 $$ and $$ x+y=4 $$ are:

1. $$ (x,y) = \left(\frac{3+\sqrt{7}}{2},\frac{5-\sqrt{7}}{2}ight) $$
2. $$ (x,y) = \left(\frac{3-\sqrt{7}}{2},\frac{5+\sqrt{7}}{2}ight) $$
The solutions are $$ \left(\frac{3+\sqrt{7}}{2},\frac{5-\sqrt{7}}{2}ight) $$ and $$ \left(\frac{3-\sqrt{7}}{2},\frac{5+\sqrt{7}}{2}ight) $$.

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