How do you find the parametric equations for the rectangular equation #x^2+y^2=16#?

Question

Aurora Alvarado · Accepted Answer

#((x), (y)) = ((4 cos t),(4 sin t))#

the most sensible/common paramaterisation here is to recognise that this is a circle, or just to acknowledge the Pythagorean identity: #cos^2 t + sin^2 t = 1#, that we could use here

so if we take your equation

#x^2+y^2=16#

...and re-write it slightly as

#(x/4)^2+(y/4)^2=1#

then we see that if we set

#x/4 = cos t# and #y/4= sin t#

we can use the identity

So the parameterisation is

#((x), (y)) = ((4 cos t),(4 sin t))#

so that, just to check, #x^2+y^2= 4^2 cos^2 t + 4^2 sin^2 t = 16#

Cameron Alexander · Answer

#{x = 4(1-m^2)/(1+m^2), y = (8m)/(1+m^2)}# for #-oo < m < oo#

This circle

#c->x^2+y^2=4^2#

is origin centered with radius #4#. Think of a line with a fixed point in #p_0={-4,0}#.

This line intersects the circle in one more point, depending on its declivity.

#l -> y - y_0 = m(x-x_0)# or
#l->y = m(x+4)#

The intersection #c nn l# is obtained solving

#{ (y-m(x+4)=0), (x^2+y^2-4^2=0) :}#

giving for #{x,y}# the solutions

#{x = -4,y=0}# the fixed point, and
#{x = 4(1-m^2)/(1+m^2), y = (8m)/(1+m^2)}#

This last solution, gives us a circle's parameterization for #-oo < m < oo#

Attached also a comparative plot for #x(m)# in blue and for #y(m)# in red. As can be observed, for #-1 le m le 1# the parameterization covers the positive #x# values and the whole #y# range. To obtain the negative #x# values we need to extend the range until #pm oo#

Aurora Alvarado · Answer

undefined To find the parametric equations for the rectangular equation $x^2 + y^2 = 16$, we can use the parametric representation of a circle. Let $x = r\cos(t)$ and $y = r\sin(t)$, where $r$ is the radius of the circle and $t$ is the parameter.

In this case, the radius of the circle is $r = 4$ (since $16 = 4^2$).

Therefore, the parametric equations for $x^2 + y^2 = 16$ are:

$$ x = 4\cos(t) $$
$$ y = 4\sin(t) $$