Understanding the Trajectory Equation: X^2+Y^2=5

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Andrei0408
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Homework Statement
Describe the motion of a mass point with a given law of motion
Relevant Equations
I've attached a picture
So I've done the first part of the problem, I just need to understand the x^2+y^2=5 part. I believe in order to get to this (I think this is called the trajectory equation, but please correct me if I'm wrong), we wrote sin(3t) = x/5 and cos(3t) = y/5, we raised each of these to the power of 2 and added them. But in the solution it says that it's the equation of a circle in the plane XOY. How do we know it's the equation of a circle?
 

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PeroK said:
Do you know the equation for a circle of radius ##R##, centred on the origin?
Not really
 
PeroK said:
That's fairly basic knowldge to be missing. It's not hard to find online.
I searched it and managed to understand but now I'm having some trouble with another exercise. I've been given the motion laws x and y and I need to find the trajectory equation, I've done the same steps that I did for last exercise, but the solution isn't right. I've attached the way I worked below, the result should be x^2+(y-2)^2=4
 

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Andrei0408 said:
I searched it and managed to understand but now I'm having some trouble with another exercise. I've been given the motion laws x and y and I need to find the trajectory equation, I've done the same steps that I did for last exercise, but the solution isn't right. I've attached the way I worked below, the result should be x^2+(y-2)^2=4
Perhaps the centre of the circle is not the origin?

In general a circle is of the form: $$(x - a)^2 + (y - b)^2 = R^2$$ where ##(a, b)## is the centre.

Can you see how to get your equation into that form?
 
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PeroK said:
Perhaps the centre of the circle is not the origin?

In general a circle is of the form: $$(x - a)^2 + (y - b)^2 = R^2$$ where ##(a, b)## is the centre.

Can you see how to get your equation into that form?
I see so in this case the centre of the circle would be C(0,2) and radius 2.
Thank you for your replies!