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Aug
5
comment Rotate a point on a circle with known radius and position
Accepted your answer since it is more detailed. Too bad that SE doesn't allow to accept multiple answers. :)
Aug
5
accepted Rotate a point on a circle with known radius and position
Aug
5
comment Rotate a point on a circle with known radius and position
Nice! Upvoted you!
Aug
5
comment Rotate a point on a circle with known radius and position
This is also helpful, but I don't understand there the radius $r$ disappeared in the final solution. Isn't it supposed to be there as well?
Aug
5
comment Rotate a point on a circle with known radius and position
Ah, sounds good! Actually, it's multiplying two matrix. Thanks a lot!
Aug
5
comment Rotate a point on a circle with known radius and position
where A,B,C are the vectors representing their respective points. -- how to build these vectors? Also, how to multiply $M(B - A)$?
Aug
5
comment Rotate a point on a circle with known radius and position
@Shailesh I saw something on Wikipedia, but didn't really understood the things. I think I know what the origin change means.
Aug
5
revised Rotate a point on a circle with known radius and position
edited title
Aug
5
asked Rotate a point on a circle with known radius and position
Aug
4
comment Move a point with known angle on a circle
Can you please add a general formula where I have as input the initial point ($x$ and $y$), the radius $R$ and the angle? Maybe it would be nice to have support for different circle centers. Is that possible?
Aug
4
comment Move a point with known angle on a circle
In this solution, where does the $R$ value appear in the formula?
Aug
4
reviewed Approve Co - ordinates of a point lying on the perpendicular bisector of a segment.
Aug
4
accepted Co - ordinates of a point lying on the perpendicular bisector of a segment.
Aug
4
asked Co - ordinates of a point lying on the perpendicular bisector of a segment.
Aug
4
awarded  Talkative
Jun
15
comment If $x ^ 6 = x$, prove that $x ^ 2 = x$, in a ring
@BillDubuque That's also interesting. Thanks! (Y)
Jun
15
revised If $x ^ 6 = x$, prove that $x ^ 2 = x$, in a ring
edited body
Jun
15
comment If $x ^ 6 = x$, prove that $x ^ 2 = x$, in a ring
@MichaelBurr Thanks! I understood the way to solve it! However, can you please explain what Simplifying the even coefficients means? Why did the coefficients disappear? Also, why $x ^ 4 = x ^ 2$ instead of $x ^ 4 = - x ^ 2$?
Jun
15
accepted If $x ^ 6 = x$, prove that $x ^ 2 = x$, in a ring
Jun
15
comment If $x ^ 6 = x$, prove that $x ^ 2 = x$, in a ring
@Mathmo123 The information about ring yes, but what I'm thinking at is could it be that it's referring to zero divisors only? I think that it would make it easier, however is there a possible solution without supposing that $x$ is a zero divisor?