# Parametric equation of the circle which starts on the $x$-axis when $t=0$

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?

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:
• @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. Mar 1 at 8:33