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

 Mar 12, 2019
 #1
avatar+233 
+2

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!

 Mar 12, 2019
 #2
avatar+21978 
+3

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 \(r\):

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

 

laugh

 Mar 12, 2019

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