Doesn't the position of unit circle affect the answer in this question? This is a KVPY-SA 2017 Question:

Section 5 - PartB-Mathematics
61)Let S be the circle in $xy$-plane which touches the x-axis at point A, the y-axis at point B and the unit circle $x^2 + y^2 = 1$ at point C externally. If O denotes the origin, then the angle OCA equals
A) $5π/8$
B) $π/2$
C) $3π/4$
D) $3π/5$

So, I'm preparing for KVPY this year and that's how I came upon this question. There's nothing wrong with the question. According to Wikipedia, It's normally placed in the origin. So I placed it and got the answer as $5π/8$. When I checked the KVPY answer key, I found my answer's right.
On googling the question (to see whether my procedure was right,)  I found a lot of solutions to this question like this one by toppr. Here's the accompanying figure:

(Please note that the axes in their image are interchanged.)
I know there are a lot of ways to find the answer but it's not the method in which they found the answer that makes me mad. In their solution, they didn't even keep the unit circle on the origin and yet they found the answer!!!!
So doesn't the position of the unit circle affect the answer.
Can anyone please explain the logic behind this?
Please forgive me if this is a dumb question. I can't just get the idea.
 A: The figure given in the toppr answer is wrong.
As a consequence, that answer also gives the wrong value of the radius of the circle.
But you don't actually need to know the radius of the circle.
All you really need to know is that the centers of both circles are on
the line $y = x$ and therefore so is the point of tangency of the circles, $C.$
So with the incorrect small circle removed from the figure from toppr, and with the $x$ and $y$ axes in the conventional orientation
($x$ axis horizontal), we have the following.
Note that the point of tangency $C$ is exactly where the toppr figure said it would be; they just used an incorrect circle to get the correct tangency point.
Using the correct circle would tell us that the distance $OC$ is $1,$
but since that information is not used in answering the question,
I have omitted the unit circle from the figure.

Now observe that $\triangle APC$ is isosceles. You know the angle at the apex $P$, so you can work out the two base angles (which must be equal).
And now that you know $\angle APC$ and $\angle CAP$
you can add them to get the exterior angle $\angle OCA$,
or you can subtract $\angle ACP$ from $\pi$ as the toppr answer does;
either way gets the correct answer.
