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# ANALYTIC GEOMETRY

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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

#3
+109497
+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
+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
+111360
+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

Dec 20, 2019
#3
+109497
+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

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

Melody Dec 20, 2019
#4
+111360
+1

CPhill  Dec 21, 2019