SOLUTION: ms. russo is hiring an architect to create blue prints for a circular pool in her fenced in backyard. the rectangular fence neasures 50*4 square feet. the pool needs to be as big a

Algebra ->  Trigonometry-basics -> SOLUTION: ms. russo is hiring an architect to create blue prints for a circular pool in her fenced in backyard. the rectangular fence neasures 50*4 square feet. the pool needs to be as big a      Log On


   



Question 827108: ms. russo is hiring an architect to create blue prints for a circular pool in her fenced in backyard. the rectangular fence neasures 50*4 square feet. the pool needs to be as big as possible within the given space. local building code states that pools have to be at least 3 feet from the fence.
a. sketch thefence and pool on the graph below
b. determine the diameter and circumference of the pool
c. if the bottom left of the fence is considered the origin on the Cartesian coordinate plane and the entire pool is in quadrant 1 then create the circle equation for the pool in circle radius form and in general form
I don't get how o find the center of the circle and the radius

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
ms. russo is hiring an architect to create blue prints for a circular pool in her fenced in backyard. the rectangular fence neasures 50*4 sq ft = 200 sq ft.
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the pool needs to be as big as possible within the given space.
Let the rectangle be a square:
Each side is sqrt(200) = 10sqrt(2) ft.
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local building code states that pools have to be at least 3 feet from the fence.
radius of the pool is (5-3) = 2 ft
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a. sketch the fence and pool on the graph below
b. determine the diameter and circumference of the pool
Diameter = 4 ft; Circumference = 4pi ft
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c. if the bottom left of the fence is considered the origin on the Cartesian coordinate plane and the entire pool is in quadrant 1 then create the circle equation for the pool in circle radius form and in general form
Center at (2,2) because radius = 2.
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Equation: (x-2)^2 + (y-2)^2 = 4
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Cheers,
Stan H.
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