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reflection of x^2 + y^2=1 with the ine x+2y=7

Guest Jan 14, 2016

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 #1
avatar+78741 
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I believe this is what you want : https://www.desmos.com/calculator/tdlapeleht

 

The circle with its center at the origin, x^2 + y^2 = 1,  is reflected across the line x + 2y = 7.

 

The reflected circle will have its center on  the line  that is perpendicular to x + 2y = 7 →  y = 2x.  The intersection of this line and x + 2y = 7  occurs at (1.4, 2.8). So.......the center of the reflected circle will be as distant from this point as the circle at the origin is from this point. And this point is 1.4 units from the right and 2.8 units up from the center of x^2 + y^2 = 1.

 

So......the reflected circle's center will be 1.4 units from the right and 2.8 units up from (1.4, 2.8)......i.e., the circle will have the equation (x - 2.8)^2 + (y - 5.6)^2  = 1

 

 

 

cool cool cool

CPhill  Jan 14, 2016
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2+0 Answers

 #1
avatar+78741 
+10
Best Answer

I believe this is what you want : https://www.desmos.com/calculator/tdlapeleht

 

The circle with its center at the origin, x^2 + y^2 = 1,  is reflected across the line x + 2y = 7.

 

The reflected circle will have its center on  the line  that is perpendicular to x + 2y = 7 →  y = 2x.  The intersection of this line and x + 2y = 7  occurs at (1.4, 2.8). So.......the center of the reflected circle will be as distant from this point as the circle at the origin is from this point. And this point is 1.4 units from the right and 2.8 units up from the center of x^2 + y^2 = 1.

 

So......the reflected circle's center will be 1.4 units from the right and 2.8 units up from (1.4, 2.8)......i.e., the circle will have the equation (x - 2.8)^2 + (y - 5.6)^2  = 1

 

 

 

cool cool cool

CPhill  Jan 14, 2016
 #2
avatar+91045 
0

Cool pic Chris :))

Melody  Jan 15, 2016

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