Let AB be a fixed line segment. Let P be a moving point such that angle APB is equal to a constant acute angle. Then point P moves along which curve?

Question

Answer is: The boundary of union of two identical intersecting circles with centers outside the common region.

What is your question? How to prove this? In that case, what have you tried? — Mees de Vries, May 07 '18 at 08:01
I don't know how to initialize this. I tried getting some idea by drawing the figure. But no use. — Priyanka, May 07 '18 at 08:08
Please use the body of the Question to give a full statement of the problem you want help with, including the context of why this problem is interesting or challenging for you. Stuffing the problem statement into the title alone may seem expeditious, but it invites confusion for Readers and relies on them to guess at what help is needed. — hardmath, May 07 '18 at 17:23

score 0 · Accepted Answer · answered May 07 '18 at 11:04

Let $P, P'$ be two such points. If $APP'B$ is cyclic, the $P'$ lies on the circumcircle of $\triangle APB$. Otherwise, the reflection of $P'$ across $\overline{AB}$, $P^*$, makes cyclic quadrilateral $APP^*B$, so $P'$ lies on the reflection of the circle across $AB$.

Let AB be a fixed line segment. Let P be a moving point such that angle APB is equal to a constant acute angle. Then point P moves along which curve?

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