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What is the area of the region enclosed by the graph of the equation \(x^2-14x+3y+70=21+11y-y^2\)   that lies below the line y=x-3?

tertre  Mar 12, 2017

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 #3
avatar+4155 
+6

I had no idea how to do this one, but I think I understand it now! surpriselaughlaugh

hectictar  Mar 12, 2017
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 #1
avatar+75316 
+5

We have

 

x^2 - 14x + 3y + 70  + y^2 - 11y - 21  = 0

 

x^2 - 14x + y^2 - 8y + 49   = 0

 

x^2 - 14x + 49 + y^2 -8y + 16  = -49 + 49 + 16

 

(x - 7)^2 + (y - 4)^2  = 16

 

This is a circle with a radius of 4  centered at (7, 4)

 

Look at the graph, here : https://www.desmos.com/calculator/9aqppc3dkl

 

The line y = x - 3    passes through (7,4)  so the line segment bounded by the circle is the diameter of that circle......so....the region of interest is just (1/2) the area of this circle  = (pi/2)*(4)^2  =

8pi units^2  ≈ 25.13 units^2

 

 

cool cool cool

CPhill  Mar 12, 2017
 #3
avatar+4155 
+6
Best Answer

I had no idea how to do this one, but I think I understand it now! surpriselaughlaugh

hectictar  Mar 12, 2017
 #2
avatar+1148 
+5

Thanks so much! smiley

tertre  Mar 12, 2017

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