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Question

# Find a parametric representation. Circle in the plane z=1 with center (3, 2) and passing through the origin.

Solution

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(I am assuming that the "origin" was meant to be the point $(0, 0, 1)$; since the circle is in the plane $z = 1$, it is impossible for it to contain the point $(0, 0, 0)$.)

Since it is in the plane $z = 1$ which is parallel to the $xy$-plane, for now we can treat the circle as if it was in the $xy$-plane. Therefore, its equation is

$(x - a)^2 + (y - b)^2 = r^2,$

where $(a,b)$ is its center, and $r$ is its radius.

Thus, from the text of the exercise, $(a,b) = (3, 2)$:

$(x-3)^2 + (y-2)^2 = r^2$

It contains the point $(0, 0)$, so we plug in $x = 0$, $y = 0$ to find $r$:

$r^2 = (0-3)^2 + (0-2)^2 = 13 \Longrightarrow \boxed{r = \sqrt{13}}$

Therefore, the full equation of this circle is

$(x-3)^2 + (y-2)^2 = 13 \tag{1}$

For the circle with the center $(a,b)$ and the radius $r$, the parametric representation is $[a+r \cos t, b + r \sin t]$. Therefore, here we have that the parametric representation is

$\boxed{\color{#4257b2}[3 + \sqrt{13} \cos t, 2 + \sqrt{13} \sin t]}$

Truly, plug this into the LHS of (1):

$(3+\sqrt{13} \cos t - 3)^2 + (2 + \sqrt{13} \sin t - 2)^2 = 13 \cos^2 t + 13 \sin^2 t = 13$

So, $(1)$ truly holds.

Finally, for now we only have $x$ and $y$ coordinates in the parametric representation, so we must remember to include $z = 1$ in the parametric representation of this circle:

$\boxed{\color{#c34632}[3 + \sqrt{13} \cos t, 2 + \sqrt{13} \sin t, 1]}$

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