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How to find a parametric equation for the circle with center at the point $(4,4)$ and radius $4$ starting on the $x$-axis when $t=0$?

Answer:

Parametric equations in vector form is

$\vec{r}$$(t)$=$(4 + 4 \sin t)$$\hat{i}$+$(4-4 \cos t)$$\hat{j}$

Can someone explain how to solve this problem?

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We know that $(x-a)^2 + (y-b)^2 = r^2$ describes a circle in $xy$-plane with center $(a,b)$. Also we know that $\sin^2 t + \cos^2 t = 1$ holds for all $t\in \mathbb{R}$. If we set $$x = a+r\sin t \\ y = b + r\cos t$$ we have $$(x-a)^2 +(y-b)^2 = r^2(\sin^2 t + \cos^2 t) = r^2$$ which is a circle. Note that because $\sin t$ and $\cos t$ are periodic with period $T = 2\pi$, it's enough to consider $t\in [0,2\pi)$. Using the mentioned parameterization, we can control the $x-$axis and $y-$axis coordinates simultaneously with the single variable $t$.

We can get the equation of an ellipse with a small modification in the equation of the circle. Using different factors for $x$ and $y$, $$x = a+r\sin t \\ y = b + r'\cos t$$ we have $$(\frac{x-a}{r})^2 + (\frac{y-b}{r'})^2 = 1$$ which is the equation of an ellipse. Parametric equations can give us really beautiful curves, for example Lissajous curve and Butterfly curve. Also take a look at this. This example shows a helix: enter image description here

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  • $\begingroup$ For parametric equation of circle , we take x=a+rcos(t), y=b+rsin(t).But here why we took x=a+rsin(t), y=b+rcos(t). $\endgroup$
    – Bee
    Mar 1 at 8:29
  • $\begingroup$ @Bee Both choices are valid. Put $t = 0, \pi/2, \pi, 3\pi/2$ and $2\pi$ in both equations to see the difference between them. $\endgroup$
    – S.H.W
    Mar 1 at 8:33

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