HELP!!

1 + $\sqrt2$ . My method isn't really orthodox and required a little bit of eyeballing it, but I'm sure it's right. Add the right side to the left, and plug it into a graphing calculator, because it's a formula for a circle. This is where the eyeballing comes in, but the largest value for x is when y equals 1. Substitute this into your equation, and you get x² + 1 -2x - 2 = x² - 2x -1. Using the quadratic formula, you get 1 - $\sqrt 2$. and 1 + $\sqrt 2$. We want the larger value for x, so the largest value of x = 1 + $\sqrt 2$ . Sorry this isn't an official method, but this is the best I could do.

Hope this helps!

Find the largest real number $x$ for which there exists a real number $y$ such that
$x^2 + y^2 = 2x + 2y$.

$\text{Let the largest real number x $ = x_{\text{max}}$} \\ \text{Let the x-center of the circle $= x_c $} \\ \text{Let the y-center of the circle $= y_c $} \\ \text{Let the radius of the circle $= r $}$

Formula:

$\begin{array}{|rcll|} \hline \mathbf{x_{\text{max}}} & \mathbf{=}& \mathbf{x_c+r} \\ \hline \end{array}$

We calculate $x_c$ and $r$:

$\begin{array}{|rcll|} \hline \mathbf{x^2 + y^2} &\mathbf{=}& \mathbf{2x + 2y} \\\\ x^2+y^2-2x-2y&=& 0 \\ x^2-2x+y^2-2y&=& 0 \\ (x-1)^2-1+(y-1)^2-1&=& 0 \\ (x\underbrace{-1}_{=-x_c})^2 +(y\underbrace{-1}_{=-y_c})^2 &=& \underbrace{2}_{=r^2} \\\\ x_c &=& 1 \\ y_c &=& 1 \\ r &=& \sqrt{2} \\\\ \mathbf{x_{\text{max}}} & \mathbf{=}& \mathbf{x_c+r} \\ \mathbf{x_{\text{max}}} & \mathbf{=}& \mathbf{1+\sqrt{2}} \\ \hline \end{array}$