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What is the area of the portion of the circle defined by \(x^2-12x+y^2=28\) that lies above the \(x\)-axis and to the right of the line \(y=6-x\)?

 Apr 19, 2019
 #1
avatar+104647 
+1

We could use either trig or calculus to solve this.....but a graphical solution seems WAY easier

 

Look at the graph here : https://www.desmos.com/calculator/cd8js1dkja

 

The line y = 6 - x   forms a diameter with its intersection of the circle

 

Notice that the line  y = x - 6 is perpendicular to y = 6- x....so.... it  forms another diameter....and these lines divide the circle into 4 equal parts

 

So....the part lying to the right of y = 6 - x  and above the x axis =  (1/4) area of circle + (1/8) area of the circle  =

 

(3/8)  area of the circle

 

The radius  =  8

 

So....the area we want is   (3/8) * pi (8)^2  =  24 pi units^2

 

 

cool cool cool

 Apr 19, 2019
 #2
avatar+10603 
+2

What is the area of the portion of the circle?

laugh

 Apr 19, 2019
edited by Omi67  Apr 19, 2019

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