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I've been thinking a lot about this question for a while now, I checked various books on how one can find the locus of something, but I just can't understand. This is not a "homework question" but a question that has been bugging me for a while now. Could anyone tell me what this question even means? It seems to be forming a strange curve which doesn't exactly resemble an arc of a circle.

Godot Roy
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    If I understand correctly the point $Q$ is obtained from $P$ by a fixed rotation around point $A$. Where do you end up when you apply a rotation around $A$ to a fixed straight line? – Michal Adamaszek Oct 31 '23 at 11:26
  • If we move P on the line, the point P moves all the way to infinity and Q and P seem to drift further and further away, but Q does not move in a straight line. It moves in a curve-like way, but really I am unable to discern anything any futher. I've drawn many diagrams. – Godot Roy Oct 31 '23 at 12:27

1 Answers1

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Draw out the figure as follows :

Begin by shifting origin to the point A.

Now, let the equation of the given straight line by $xcos(a) + ysin(a) = p$,

(here a & p are fixed clearly)

i.e. perendicular distance from A to the line = P

Now take any point P on this line such that :

it makes some angle '$t$' with the shortest distance from A to $xcos(a) + ysin(a) = p$.

Thus, our point Q lies on circle with radius $p.sec(t)$.

Assume some fixed angle (in counter-clockwise sense) that AQ makes with AP.

Call it '$b$'.

$\implies Q $ = $((p.sec(t)cos(a+b+t)$) , $(p.sec(t)sin(a+b+t)))$ = $(x,y)$ ;

$\implies x^2 + y^2$ = $(p.sec(t))^2$ ----------------(1)

$\implies x = p(sin(a+b)sec(t) - cos(a+b))$ ;

$\implies \displaystyle\frac{x+ p.cos(a+b)}{p.sin(a+b)}$ = $sec(t)$ ----------(2)

So, using (1) & (2), we wet the desired locus of Q ,i.e.

($(x+h)^2 + (y+k)^2$)($sin(a+b))^2$ = $((x+h)$ + $pcos(a+b))^2$

Note that here : $A = (h,k)$.

Akshat Mishra
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