How to shift a function of a half circle on the x-axis?

Question

I want to have a half circle function from $x=0$ to $x=0.056$ with $f(0)=0$ and $f(0.056)=0$, and the highest point at $f(0.056/2) = 0.001$

I tried: $$ f(x) = \sqrt{0.001^2 - ( ( [0.001\times0.056]-x ) / 0.056)^2 } $$ but it will give me an error, because the number in the sqrt gets negative.

$f(x)=0.001\sqrt{1-\dfrac{(x-0.028)^2}{0.028^2}}$ is a half ellipse with the desired points — J. W. Tanner, Nov 05 '19 at 22:04
@J.W.Tanner No, that equation does not work. It gives me complex values. :( — thale00, Nov 05 '19 at 22:09
If $0\le x\le 0.056$, then $-0.028\le x-0.028\le 0.028$, so $(x-0.028)^2\le0.028^2$, so $\dfrac{(x-0.028)^2}{0.028^2}\le1$ so $1-\dfrac{(x-0.028)^2}{0.028^2}\ge0$ so $f(x)$ is real — J. W. Tanner, Nov 05 '19 at 22:14
@J.W.Tanner : OP is looking for a circle. Its center is not on the x-axis because the height is not equal to half the distance between x-intercepts. — MPW, Nov 05 '19 at 22:18
Possible duplicate of Get the equation of a circle when given 3 points — MPW, Nov 05 '19 at 22:19
@J.W.Tanner I put f(x) = 0.001*sqrt(1 - (x-0.028)^2 / (0.028)^2 ) in wolframAlpha or in Matlab and it says, that there are complex values... — thale00, Nov 05 '19 at 22:23
I was confused because OP wanted "a function from $x=0$ to $x=0.056$" — J. W. Tanner, Nov 05 '19 at 23:31
@J.W.Tanner : I understand, I was thinking along the same lines and nearly posted a wrong answer. Then I noticed the height was wrong. Since a circle was specified, I realized the two endpoints do not lie in a diameter. — MPW, Nov 06 '19 at 01:15

score 1 · Answer 1 · answered Nov 05 '19 at 22:13

Hint: Suppose the center is at $(a,b)$ and the radius is $r$, so that the equation of the circle is of the form $$(x-a)^2 + (y-b)^2 = r^2.$$ Since you know three points on the circle, you can write a system of three equations in three unknowns. Solve that system for the unknown parameters $a$, $b$, and $r$. Your function will be $$f(x)=b + \sqrt{r^2 - (x-a)^2}$$ This function has domain $[a-r, a+r]$, the projection of the horizontal diameter onto the $x$-axis.

How to shift a function of a half circle on the x-axis?

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