How can you find the locus of a point that is the same distance from a exponential function? Okay so this is a question that I came up with few weeks ago, and I still cannot find a answer to this problem.
Abstract idea: If you draw a line that is perpendicular to the tangent line of a point $(a,2^a), a∈R$ and have point A that is on the tangent line with a distance of one from the previous point, would it be possible to find a locus of A?
Here is my question better worded:
Let $f(x)=2^x$.  Point $A(p,q)$  satisfies the following:

*

*A is a point on function $g(x)=-\frac{1}{f'(a)}(x-a)+f(a),(a∈R,p≥a)$

*A is a point on circle $C: (x-a)^2+(y-f(a))^2=1$
Find the locus of A.
Picture for better understanding
I can express $p$ and $q$ in $a$, and tried to place $p$ and $q$ in $x$ and $y$, respectively, but failed to get an answer. Now I cannot think of a way to solve this problem from that point.
Is finding the locus of A possible?
Also, I'm just a high school student, so if this question is too obvious on a high school level, feel free to criticise my naivety.
 A: How about this:

*

*The point $A$ is on the line $ y = g(x) $, thus
$$ q = g(p) = -\frac{p - a}{f'(a)} + f(a) \tag{1}\label{eq1} $$


*The point $A$ is on the circle $ C: (x - a)^2 + (y - f(a))^2 = 1 $, thus
$$ (p - a)^2 + (q - f(a))^2 = 1 \tag{2}\label{eq2} $$
From \eqref{eq1}, you have $ q - f(a) = -\frac{p - a}{f'(a)} $.
Plugging that into \eqref{eq2}, you get
$$ (p - a)^2 + \frac{(p - a)^2}{(f'(a))^2} =
   (p - a)^2 \left( 1 + \frac{1}{(f'(a))^2} \right) = 1 $$
From which you can easily isolate $p$:
$$ p = a \pm \frac{f'(a)}{\sqrt{1 + (f'(a))^2}} $$
Note that since $ f'(a) = 2^a \ln 2 > 0 $, the condition $ p \geq a $ implies that you should only keep the $+$ sign, i.e. the "rightmost" intersection of the line
$ y = g(x) $ with the circle $C$.
So, plugging $ p - a = \frac{f'(a)}{\sqrt{1 + (f'(a))^2}} $ back into \eqref{eq1},
you finally get the locus of $ A $:
$$ A(p, q) = \left( a + \frac{f'(a)}{\sqrt{1 + (f'(a))^2}},
                    f(a) - \frac{1}{\sqrt{1 + (f'(a))^2}} \right) $$
