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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?

 Dec 20, 2019

Best Answer 

 #3
avatar+106915 
+2

Thanks Chris, I think you overlooked one of the conditions.

 

I just did this in ful but it was lost.

 

Anyway here is my pic

 

\(Area = \frac{135}{360}\times \pi\times 6^2\\ Area = \frac{27\pi }{2} \; units\; squared \)

 

 Dec 20, 2019
 #1
avatar
+1

The circle has a radius of 12, and the line y = 6 - x splits it in half, so the area is 1/2*12^2*pi = 72*pi.

 Dec 20, 2019
 #2
avatar+106519 
+2

See the graph  here :    https://www.desmos.com/calculator/qeidqnvtts

 

The intersection of this line and the circle  will form the diameter of the  circle....to find the radius we can  complete the square  on x  in the equation of the circle......so we have....

 

x^2 - 12x  +36  + y^2   = 28 + 36

 

(x + 6)^2  + y^2 =  64

 

The center is  (6, 0)

 

The radius  = sqrt (64)   = 8

 

So....the area we are looking for  is

 

(1/2) pi (8)^2 =   

 

32 pi units^2

 

 

 

cool cool cool

 Dec 20, 2019
 #3
avatar+106915 
+2
Best Answer

Thanks Chris, I think you overlooked one of the conditions.

 

I just did this in ful but it was lost.

 

Anyway here is my pic

 

\(Area = \frac{135}{360}\times \pi\times 6^2\\ Area = \frac{27\pi }{2} \; units\; squared \)

 

Melody Dec 20, 2019
 #4
avatar+106519 
+1

Ah...thanks, Melody....I forgot about that.....!!!!

 

 

cool cool cool

CPhill  Dec 21, 2019

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