So I know the formula for a semi circle is

$$y = \sqrt{r^2 - x^2}$$

However, what if I wanted to find the equation for a semi circle who's diameter is at the top of the graph?

Would this be the best solution?

$$y = r-\sqrt{r^2 - x^2}$$


  • $\begingroup$ Not sure what you are asking for. The result of what you are asking for is here: wolframalpha.com/input/… $\endgroup$ – gt6989b Feb 25 '14 at 21:44

The $r$ at the start is optional. It just shifts the semicircle up so the bottom is tangent to the $x$ axis. If you delete it the semicircle is below the $x$ axis. In both cases it has the orientation you desire. Good work.


It also can be $y=-\sqrt{r^2-x^2}$

  • $\begingroup$ Sorry, should have specified that I need the function to start on the x-axis. $\endgroup$ – dkind Feb 25 '14 at 22:01
  • $\begingroup$ One way to see this is that the equation for the full circle, $ \ x^2 + y^2 = r^2 \ $ does not describe a single function (as it fails the "vertical line test"), but rather represents two separate functions, since $ \ y = \pm \sqrt{r^2 - x^2} \ , $ with the positive square-root corresponding to the curve for the "upper semi-circle" and the negative square-root, the "lower semi-circle". $\endgroup$ – colormegone Feb 25 '14 at 23:45

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