How do I find the equation for a semicircle with a radius of 2 on the x axis?
I'm doing a calculus project where we have to make a model of some graph rotated about the y axis. I am doing a fish bowl and I have most of it understood and ready. The only thing I'm not sure of is what the equation of the semicircle is if its center is at (0,1.5). I added a picture to help illustrate it. Thank you for your help. Image of semicircle
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$\begingroup$Equation of a semi-circle at the origin: $y= (r^2-x^2)^{1/2}$
To translate this to centre (a, b): $y = \sqrt{r^2 - (x - a)^2} + b$
The solution for your example SHOULD be: $y=\sqrt{4-x^2}+1.5$
Except, it's not. The image attached shows this graph, but rotated 90 degrees clockwise. To resolve this, I'm going to perform a refection in the line y=x. This can be done by swapping the x and y values as shown: $x=\sqrt{4-y^2}+1.5$
$\endgroup$ 0 $\begingroup$In general, if a circle has center $(a,b)$ and radius $r$, then its equation is $$(x-a)^2 + (y-b)^2 = r^2.$$
The equation for that semicircle is therefore $$x^2+(y-1.5)^2 = 4,$$ with the restriction $x\geq 0$. If you wish, you can rewrite this as $$x = \sqrt{4-(y-1.5)^2},$$ where $y\in [-\frac{1}{2},3\frac{1}{2}]$.
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