Question 218172
equation of the circle is x^2 + y^2 = 10
solve for y:
y^2 = 10-x^2
y = +/- sqrt(10-x^2))
You want the upper half of the circle only, so you want:
y = sqrt(10-x^2))
You will reject y = -sqrt(10-x^2) because that would give you the bottom half of the circle which you don't want.
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Your domain is determined by sqrt(10-x^2))
Your domain has to result in real values of y in your range.
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If sqrt(10-x^2) is positive or 0, then your equation is valid because that would result in real values of y in your range.
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If sqrt(10-x^2) is negative, then your equation is invalid because you would have the square root of a negative number which would not result in real values of y in your range.
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You need to find out when 10-x^2 is >= 0
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add x^2 to both sides of this equation to get:
10 >= x^2
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multiply both sides of this equation by -1 to get:
-10 <= -x^2
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add x^2 to both sides of this equation and add 10 to both sides of this equation to get:
x^2 <= 10
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Your domain will result in real values of y in your range when x^2 <= 10.
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x^2 is <= 10 when x < sqrt(10) and when -x > - sqrt(10)
Let x = sqrt(10), then x^2 = 10
Let x = -sqrt(10), then x^2 = 10
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Your domain is when -sqrt(10) <= x <= sqrt(10)
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A graph of your equation is shown below:
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{{{graph(300,300,-5,5,-5,5,sqrt(10-x^2))}}}
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since sqrt(10) = +/- 3.16227766, the graph looks accurate as far as the naked eye can see.
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