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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 ?

ant101  Jan 6, 2018
 #1
avatar+88836 
+2

Rearrange as

 

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

 

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

 

(x - 3)^2  + (y  - 7)^2  =    25

 

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

 

The part lying below y  = 7 is the  area of 1/2 of this circle  =

 

pi * 5^2  / 2   =

 

(25/2)pi ≈  39.3  units^2

 

 

cool cool cool

CPhill  Jan 6, 2018
 #2
avatar+553 
+1

check again

ant101  Jan 6, 2018

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