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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

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

What is the area of the portion of the circle?


 Apr 19, 2019
edited by Omi67  Apr 19, 2019

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