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What is the area of the circle defined by \(x^2-6x +y^2-14y +33=0\)  that lies beneath the line \( y=7\) ?

tertre  Mar 14, 2017
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 #1
avatar+76972 
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x^2 - 6x + y^2 - 14y + 33 = 0   

 

x^2 - 6x + y^2 - 14y   = -33       complete the square on x and y

 

x^2 - 6x + 9 + y^2 - 14y + 49   =  -33 + 9 + 49    factor and simplify

 

(x - 3)^2  + (y - 7)^2   =   25

 

This a circle that is centered at (3, 7 )   with a radius of 5

 

So.....the area of the portion of the circle that lies below y = 7 is the area of the half circle

 

And this is  =   (1/2)* pi (5^2)  =  12.5 pi  units^2  ≈  39.27 units^2

 

 

cool cool cool

CPhill  Mar 14, 2017
 #2
avatar+1167 
0

Yes! It's correct!

tertre  Mar 14, 2017

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