Motion along a circle The problem below is from the textbook I'm reading from. I'm also stating the solution to the problem after the problem statement. Someone kindly verify the legitimacy of my solution.
Problem Statement:

Show that the vector valued function  $$
 \textbf{r}(t)=2\textbf{i}+2\textbf{j}+\textbf{k}+\cos
 t\left(\frac{1}{\sqrt{2}}\textbf{i}-\frac{1}{\sqrt{2}}\textbf{j}\right)+\sin
 t\left(\frac{1}{\sqrt{3}}\textbf{i}+\frac{1}{\sqrt{3}}\textbf{j}+\frac{1}{\sqrt{3}}\textbf{k}\right)
 $$ describes the motion of a particle moving in the circle of radius
$1$, centered at $(2,2,1)$.

My Solution:
Part 1: Proving $\textbf{r}(t)$ represents a circular path.
Let us for now assume this function indeed defines a circular path and let the center be $(a,b,c)$. Let $\textbf{R}(t)$ be the radius vector from the center to any point on the curve $\textbf{r}(t)$. Then
$$
\textbf{R}(t)=\textbf{r}(t)-\langle a,b,c \rangle=\left\langle 2-a+\frac{\cos t}{\sqrt{2}}+\frac{\sin t}{\sqrt{3}}, 2-b-\frac{\cos t}{\sqrt{2}}+\frac{\sin t}{\sqrt{3}},1-c+\frac{\sin t}{\sqrt{3}} \right\rangle
$$
The tangent vector to $\textbf{r}(t)$ is obtained by differentiating $\textbf{r}(t)$. Let it be $\textbf{v}(t)$.
$$
\textbf{v}(t)=\textbf{r}'(t)=\left\langle -\frac{\sin t}{\sqrt{2}}+\frac{\cos t}{\sqrt{3}}, \frac{\sin t}{\sqrt{2}}+\frac{\cos t}{\sqrt{3}},\frac{\cos t}{\sqrt{3}} \right\rangle
$$
If $\textbf{r}(t)$ represents a circular path then $\textbf{R}(t)\cdot\textbf{v}(t)=0$. This gives us
$$
\textbf{R}(t)\cdot\textbf{v}(t)=\left(\frac{a-b}{\sqrt{2}}\right)\sin t + \left(\frac{5-a-b-c}{\sqrt{3}}\right)\cos t=0
$$
only if
$$
a=b \text{ & } a+b+c=5
$$
We can see that $(a,b,c)$ exist that can satisfy this condition and hence $\textbf{r}(t)$ represents a circular path.
Part 2: Finding the center.
Since $\textbf{r}(t)$ is a circular locus, the radius vector's magnitude would be constant, i.e $|\textbf{R}(t)|=k$. This means
$$
\frac{d}{dt}|\textbf{R}(t)|=0 \implies \frac{d}{dt}|\textbf{R}(t)|^2=0
$$
Doing differentiation and some algebra gives us
$$
|\textbf{R}(t)|\frac{d}{dt}|\textbf{R}(t)|=\left(\frac{a-b}{\sqrt{2}}\right)\sin t+\left(\frac{4-a-b}{\sqrt{3}}\right)\cos t=0
$$
From part 1, we know that $a=b$ and $a+b+c=5$. Using this we get
$$
\left(\frac{4-2a}{\sqrt{3}}\right)\cos t=0 \implies a=2 \implies b=2 \implies c=1.
$$
Part 3: Finding radius
Substituting values of $a$, $b$ and $c$ into the expression for $|\textbf{R}(t)|$ gives $|\textbf{R}(t)|=1$. I've spared the algebra. This completes my solution/proof.
My main question
In Part 1, I have assumed $\textbf{r}(t)$ to be a circular path and found that real numbers $a$, $b$ and $c$ exist that satisfy the condition for $\textbf{r}(t)$ to be circle: that radius is perpendicular to tangent. If $\textbf{r}(t)$ wasn't a circle, then $a$ and/or $b$ and/or $c$ wouldn't exist. Is this argument watertight? I think so because there is no other curve that satisfies this condition, at least as far as I know.
 A: The most straightforward proof computes $$
\left| \textbf{r}(t)-(2\textbf{i}+2\textbf{j}+\textbf{k})\right|,$$ shows, via $$\left| \cos t\left(\frac{1}{\sqrt{2}}\textbf{i}-\frac{1}{\sqrt{2}}\textbf{j}\right)+\sin t\left(\frac{1}{\sqrt{3}}\textbf{i}+\frac{1}{\sqrt{3}}\textbf{j}+\frac{1}{\sqrt{3}}\textbf{k}\right)\right|,$$ that it equals $1,$ then finally argues that since the vector's path is at a fixed $1$-unit distance away from $(2,2,1),$ the path is by definition a circle with centre $(2,2,1)$ and radius $1.$
Your current approach is invalid because all you have proven is that if the path is circular, then it has centre $(2,2,1)$ and radius $1.$
To repair your proof without adopting mine, I suggest either completing it by showing/justifying that the path is actually circular, or,  restructuring and shortening it by using the given point $(2,2,1)$ instead of finding it.

Addendum

I still have a question though: why is my solution invalid? I made an assumption and proved it right. This isn't incorrect IMO.

To be clear: the task is not to show that the path is a particular circle (say, Circle Camille) among all possible circles.
The task is to show that the path is Circle Camille, as opposed to Ellipse Elsa, Hexagon Hzael, Triangle Teddy, etc.
Your mistake is closely related to circular reasoning.
