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I was working through my Precalculus 12 book, when I came across these questions:

Is each point on the unit circle? Give evidence to support your answer

a) $(0.65, -0.76)$

b) $\left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$

My book says that both of these points lie on the unit circle, but I can't understand how.

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    $\begingroup$ By definition of unit circle, any point $(x,y)$ of $\mathbb R^2$ is in the unit circle if, and only if, $\sqrt{x^2+y^2}=1$ or equivalently $x^2+y^2=1$. The first point isn't in the unit circle. It's easy to realise this because the denominator of $\dfrac{65^2+76^2}{100^2}$ is odd. $\endgroup$
    – Git Gud
    Dec 31, 2014 at 19:15
  • $\begingroup$ How far away from the origin are all points on the unit circle? Are both of these points this far away? $\endgroup$
    – beanshadow
    Dec 31, 2014 at 19:15
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    $\begingroup$ @GitGud: An odd $100^2$ is odd indeed! (I assume that you meant that the numerator is odd.) $\endgroup$ Dec 31, 2014 at 19:25
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    $\begingroup$ @BrianM.Scott Ahaha. Yes, thank you. $\endgroup$
    – Git Gud
    Dec 31, 2014 at 19:27

3 Answers 3

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Hint:

A point with coordinates $(a,b)$ is in the unit circle if and only if

$$ a^2+b^2=1. $$

Explanation: The unit circle is by definition a circle with radius equal to $1$ and center $(0,0)$, so the distance of the points $(x,y)$ in that circle to the center is equal to $1$, hence by the distance formula :

$$ \sqrt{a^2+b^2}=1\iff a^2+b^2=1. $$

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If you are studying the unit circle, then b) should be a familiar cartesian coordinate, as it equivalent to the polar coordinate $\left(1,\frac{5\pi}{4}\right)$. To determine if a) is on the unit circle, you can do as others have suggested, and check the value of $$0.65^2+(-0.76)^2$$ If it equals $1$, it is on the unit circle. It does not equal $1$, which is easy to see by using a calculator. So technically that point is not on the unit circle, although the value is so close to $1$ that it is reasonable to assume the book rounded the decimals. I would make a note of that if I were turning this in for homework.

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we have $x^2+y^2=1$ then we plug the coordinates in this equation $$0.65^2+0.76^2=1.0001$$ and further $$\left(\frac{\sqrt{2}}{2}\right)^2+\left(\frac{\sqrt{2}}{2}\right)^2=1/2+1/2=1$$ the second point is on the unit circle

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