<s>[INST] A point $(a, b)$ on a circle with an equation $x^2+y^2=1$ defines another point $(x_1,y_1)$. What is the equation of the curve and is it a circle? The other point $(x_1, y_1)$ is defined as a point that lies on the tangent of a circle to the point $(a, b)$ and is $2$ units away from the point $(a, b)$. The question is, what is the equation of a curve the collection of all possible points $(x_1, y_1)$ trace? It seems to be a circle, but I don't understand how.
You can imagine this curve by getting two sticks that are connected at a 90 degree angle to one another, one lenght 1 and the other length 2, and spinning the lenght 1 stick in a circle around the end that isn't connected to the 2 stick, and then seeing what kind of curve the 2 stick's end that isn't connected to the 1 stick traces.
The equation of a tangent line is $y=\frac{a^2-ax+b^2}{b}$, and the distance of 2 units is given by the circle with radius 2 from the point $(a, b)$ or $(x-a)^2+(y-b)^2=4$. There are 2 intersections at points $$x=\frac{a^3\pm2\sqrt{b^2(a^2+b^2)}+ab^2}{a^2+b^2}$$
$$y=\frac{a^2b^2\pm2a\sqrt{b^2(a^2+b^2)}+b^4}{b(a^2+b^2)}$$
Knowing that $a^2+b^2=1$, substituting this into our equations we get the folowing parametric equation for this curve:
$$x=a\pm2\sqrt{1-a^2}$$
$$y=\frac{\pm(a^2\pm2\sqrt{1-a^2}a+1)}{\sqrt{1-a^2}}$$
$$-1<=a<=1$$
How do I prove that this is a circle, if it even is one? Also, is there a faster and better solution to understanding why it is a circle, if it is one?
 [/INST] 
The equation of a tangent line is $y=\frac{a^2-ax+b^2}{b}$, and the distance of 2 units is given by the circle with radius 2 from the point $(a, b)$ or $(x-a)^2+(y-b)^2=4$.

Expand the squares, rearrange the equations, and use that $\,a^2+b^2=1\,$:
$$
\begin{align}
\begin{cases}
ax + by &= a^2 + b^2 \\
x^2+y^2 -2(ax+by) + a^2+b^2 &= 4
\end{cases}
 \\[5px] \iff\quad
\begin{cases}
ax + by &= 1 \\
\color{blue}{x^2+y^2} &= 2(ax+by) + 3 = 2 \cdot 1 + 3 \color{blue}{=5}
\end{cases}
\end{align}
$$
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