Asked by rmz
Garden Area Problem. A designer created a garden
from two concentric circles whose equations are as follows:
(x+2)^2+(y-6)^2=16 and (x+2)^2+(y-6)^2=81
The area between the circles will be covered with grass. What is the area of that section?
How do you do this?
from two concentric circles whose equations are as follows:
(x+2)^2+(y-6)^2=16 and (x+2)^2+(y-6)^2=81
The area between the circles will be covered with grass. What is the area of that section?
How do you do this?
Answers
Answered by
MathMate
The equation of a standard circle with centre at (a,b) and radius r is
(x-a)² + (y-b)² = r²
By inspection of the given circle,
(x+2)^2+(y-6)^2=16 and (x+2)^2+(y-6)^2=81
we conclude that both have centres at (-2,6), therefore they are concentric.
The radii of the circles are √16=4 and √81=9.
The area between the two circles are therefore
πr1²-πr2²
=π(9²-4²)
=65π
(x-a)² + (y-b)² = r²
By inspection of the given circle,
(x+2)^2+(y-6)^2=16 and (x+2)^2+(y-6)^2=81
we conclude that both have centres at (-2,6), therefore they are concentric.
The radii of the circles are √16=4 and √81=9.
The area between the two circles are therefore
πr1²-πr2²
=π(9²-4²)
=65π
Answered by
rmz
do you have any sites that explain this as well?
Answered by
MathMate
The subject should be in any pre-calc textbook. Alternatively, you can google
"equation of a circle".
"equation of a circle".
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