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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:

enter image description here

(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.

lee
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    Umm...isnt the equation of the unit circle given? In coordinate geometry, when the eqn of a curve is given, you usually consider its position as given too. – Ishraaq Parvez Jan 04 '21 at 11:55
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    Well, the position of the unit circle is told. The equation is $x^2 +y^2 =1$ (which can be rewritten as $(x-0)^2 + (y-0)^2 = 1$), which says that the centre of that circle is exactly at $(0,0)$. – Matti P. Jan 04 '21 at 11:55
  • Then what about the answer by toppr – lee Jan 04 '21 at 11:57
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    The only way that the center would be ambiguous is if they had given the equation $(x-a)^2 +(y-b)^2 = 1$, where not both $a,b$ are given. – Vishu Jan 04 '21 at 11:57
  • I think that the circle $S$ is in your mind rather than the unit circle. – PNDas Jan 04 '21 at 11:59
  • No, I'm talking about the unit circle. You can see it's not placed at the origin in the link that I gave. – lee Jan 04 '21 at 12:02
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    @lee Answers by toppr are not of good quality. They have many wrong answers –  Jan 04 '21 at 12:04
  • But still, how did they get that answer correct. – lee Jan 04 '21 at 12:05
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    Coincidence? ${}{}$ – Vishu Jan 04 '21 at 12:13
  • Are you sure that it is coincidence. – lee Jan 04 '21 at 12:19
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    @lee: The consequence of the toppr answerer's error is that point $C$ is the wrong distance from the origin. But, luckily, $C$ is still on the line $y=x$, so all the error does is give the figure the wrong scale; the figure is still proportional to the correct one, and therefore the incorrect $\angle OCA$ will match the correct one. You'd get the same result if you took $C$ to be any (first quadrant) point you like on the line $y=x$. – Blue Jan 04 '21 at 12:29
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    @lee: Why remove the figure? It's better to include it here rather than make readers click-through to a separate site to understand what the issue is. – Blue Jan 04 '21 at 12:49
  • Added it, but that image has a problem. It has the x-axis and the y-axis interchanged. – lee Jan 04 '21 at 12:56
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    @lee: Again, it's lucky that the figure is symmetric about the line $y=x$, so that the interchanged axes (or, probably more accurately, the interchanged points $A$ and $B$) have no impact on the solution itself. I didn't even notice the mistake, but you could always just add a remark in your question that the figure has this additional flaw. – Blue Jan 04 '21 at 13:10

1 Answers1

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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.

enter image description here

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.

David K
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  • Thanks for the answer. But I would like to recommend a few changes. The figure from toppr has the axes interchanged (which you haven't corrected). And you haven't shown the position of the unit circle. And my question was: Why the answer is correct eventhough the unit circle was misplaced? which you didn't answer unfortunately. – lee Jan 04 '21 at 15:23
  • And thank you very much for this piece of advice: "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." I never thought of that. – lee Jan 04 '21 at 15:24
  • @lee I see what you mean about the vertical x axis. I should just have flipped the figure the right way around (really not that much effort as it turns out) to reduce the possibility of confusion. I've done that now. – David K Jan 05 '21 at 04:03