# Function in Y semi-circle

This is NOT a homework, I am over homework stage ages ago :-)

What function draws semicircle? $$y=f(x) = \text{something}$$ and I end up with a unit circle.

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## 2 Answers

The only "unit semicircles" that can be expressed as $y=f(x)$ are the upper half of the unit circle and the lower half (any other half circle fails the "vertical line test", so it cannot be expressed as an explicit function of $x$). Since the unit circle consists of all points $(x,y)$ that satisfy $x^2+y^2=1$, solving for $y$ gives you that $y=\sqrt{1-x^2}$ is the upper unit semicircle, and $y=-\sqrt{1-x^2}$ is the lower unit semicircle.

If you are willing to use parametric equations, you can cover more possibilities. The unit circle can be parametrized as $$\begin{array}{rcl} x & = & \cos t,\\ y & = & \sin t; \end{array}\qquad 0\leq t\leq 2\pi;$$ this "traces" the unit circle exactly once, counterclockwise, starting at $0$ radians (the point $(1,0)$). If you want the unit semicircle that starts at $\theta$ radians from the positive $x$-axis (and ends at $\theta+\pi$ radians), then you can obtain it parametrically by $x=\cos(t+\theta)$ and $y=\sin(t+\theta)$, $0\leq t\leq \pi$ (or simply $x=\cos t$, $y=\sin t$ with $\theta\leq t\leq \theta+\pi$).

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On the interval $[-1,1]$ you can use the function $\sqrt{1-x^2}$. Using the Pythagorean theorem you can see that the graph of this function is the closed upper semicircle with radius 1 and centre (0,0).

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