Answer to Question #51273 in Algebra for tinn

Question

-1≤x≤0 does the following equation x= - sqrt (1-y^2) satisfies the above condition?
i think it should be x= ±sqrt (1-y^2)for this condition

Answer 1

Equation x= - sqrt (1-y^2) determines the part of circle, because after raising both sides to the square obtain x^2+y^2=1. It is known that range of square root sqrt() is non-negative real numbers. If -1≤x≤0, then x is not positive, therefore
x=sqrt (1-y^2) does not satisfy the condition -1≤x≤0, but x=- sqrt (1-y^2) satisfies the condition -1≤x≤0. It should be
x=sqrt (1-y^2) for this condition.