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 $t=0$?


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?


1 Answer 1


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.