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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 '17 at 16:57
  • $\begingroup$ $x^2+y^2 = r^2$ for the particular circle that you exhibit. $\endgroup$ – Michael Hardy Nov 13 '17 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. $\endgroup$ – Michael Hardy Nov 13 '17 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 $\endgroup$ – Hamza Erdem Nov 13 '17 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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