# how can I draw half circle in matlab ?

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Lee on 20 Sep 2020
Edited: Lee on 21 Sep 2020
Hi.
I want to know how to draw half circle in rectangular coordinate .
(x - a)^2 + (y - b)^2 = r^2 ( y <= - (2/3)x )
I already know that in polar coordinate . ( when I using radius, theta )
like this
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xCenter_2 = 3/(2*sqrt(13));
yCenter_2 = -1/sqrt(13);
theta = 0-2/3 : 0.01 : pi-2/3;
x = radius * cos(theta) + xCenter_2;
y = radius * sin(theta) + yCenter_2;
plot(x, y), hold on;
axis square;
------
Just I want to know how to draw it " NOT use radius and theta "
I tried to this one but I failed.
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[X,Y] = meshgrid(-4:0.01:4,-4:0.01:4); % Generate domain.
Z = X.^2 + Y.^2;
contour(X,Y,Z,[1 1]);
axis square;
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John D'Errico on 20 Sep 2020
Edited: John D'Errico on 20 Sep 2020
Sigh. You DID make an effort. It is not even a terrible solution, though what you did will not work, since it will at best be a full circle, and even then, it will only be a polygonal solution. That is, a contour plot is just a piecwise linear approximation. Finally, you needed to force the contour plotter to plot ONLY a specific contour, not a complete contour plot.
The contour plot you have generated is that of a paraboloid of revolution. So not really a good choice, because it is still the full circle.
How would I solve this problem? I can probably think of a couple of solutions given some time, though it seems absolutely silly to not just use pollar coordinates, since that is the direct and obvious solution. But can I do it? I imagine so.
It is simple enough to plot the full circle.
xCenter_2 = 3/(2*sqrt(13));
yCenter_2 = -1/sqrt(13);
fxy = @(x,y) (x - xCenter_2).^2 + (y - yCenter_2).^2 - radius.^2;
fimplicit(fxy)
axis equal
But then to get only a half circle will take more work. A hack like this should work.
fxy = @(x,y) (x - xCenter_2).^2 + (y - yCenter_2).^2 - radius.^2 + (y > -(2/3)*x)*1e16;
fimplicit(fxy)
axis equal
I imagine I could do better, but why?
Or, I might formulate the problem as a differential equation, then use an ODE solver like ODE23s to solve it. If we differentiate the circle equation, it reduces to
2*(x-a) + 2*(y - b)*y' = 0
Therefore we would have
y' = -(x - a)/(y - b)
Start the solver off at one end of the arc, with initial conditions set to move in the correct direction around the circle. A stiff solver will probably be necessary, due to the singularity at y == b.
Finally, I could probably create the curve as a spline. I'm sure I could find other ways to build it. But why in the name of god and little green apples is there a reason to do so?
Just use polar coordinates.

Lee on 20 Sep 2020
Hi guy
Thank you for your detailed explanation. I gratify my curiosity.
It was just trying . I will never solve the problem like this .
Ummmmmmmmmmmmmmm ...
I'm so sorry but can you get rid of the straight line in half circle ?
John D'Errico on 20 Sep 2020
Oh. I thought you wanted that. :)
How about a conformal mapping, of sorts?
Given three points that would reperesent two sides of a square. You can use linear interpolation to find any point between them. Then just remap the points to lie at a uniform distance from the center of the circle. Honestly, that is just working in polar coordinates, while obfuscating the distinction.
Lee on 21 Sep 2020
It's too hard ;-(
But I appreciate you.
Thank you so much. Have a nice day :)~