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\begin{align} x & = r\cos t \\ y & = r\sin t \end{align}

These are parametric equations of a circle.

How can we write an equation which is non-parametric for a circle?

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    $\begingroup$ $x=x_c+r\cos t;\;y=y_c=r\sin t$ are parametric equations of a circle. Rectangular equation is $$(x-x_c)^2+(y-y_c)^2=r^2$$ Where $(x_c,y_c)$ are the coordinates of the centre $\endgroup$
    – Raffaele
    Nov 13, 2017 at 16:57
  • $\begingroup$ $x^2+y^2 = r^2$ for the particular circle that you exhibit. $\endgroup$ Nov 13, 2017 at 16:57
  • $\begingroup$ Perhaps we should note that if $t$ varies with $r$ fixed, this is a circle, but if $r$ varies with $t$ fixed, this is a straight line through the origin. $\endgroup$ Nov 13, 2017 at 17:00

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$(x-h)^2 + (y-k)^2 = r^2$, where $r$ is the radius of the circle and the center being at $(h,k)$.

The equation for the unit circle with the center in the origin would simply be $x^2 + y^2 = 1$

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it is $$x^2+y^2=r^2((\sin(t))^2+(\cos(t))^2)=r^2$$ since $$\sin(t)^2+\cos(t)^2=1$$

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  • $\begingroup$ thanks for your help $\endgroup$ Nov 13, 2017 at 18:19
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Well, consider this, suppose you want to write equation of the circle,

What do you know about the circle?

Ummmm...... Distance of all the points are equal and equal to radius?

That's enough, using that, let's derive the equation!

Let the centre be $$(x_0,y_0)$$ and radius be $$r$$ Now distance of point $(x,y)$ from centre is equal to $r$,

Using the distance formula, $$\sqrt{(x-x_0)^2+(y-y_0)^2}=r$$ Squaring, $$(x-x_0)^2+(y-y_0)^2=r^2$$ And TADA!!

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