Given an angle and a point, construct a circle through the point and tangent to the sides of the angle

Question

Given an angle $\angle ABC$ and a point $P$ inside of it, draw a circle which passed through $P$ and is tangent to both $\overleftrightarrow{AB}$ and $\overleftrightarrow{BC}$.

Well, the center has to lie on the angle bisector (obvious). The center must also lie on the perpendicular bisector of $P$ and a point $E$ on $\overleftrightarrow{AB}$ or $\overleftrightarrow{BD}$. Additionally, if $D$ is the center, then $\angle DEB = 90^\circ−\angle ABC$; in other words, $E$ is the intersection of the circle with diameter $BE$ and $AB$.

I've had little experience dealing with construction problems, so what are some "obvious" things to see?

https://www.geogebra.org/geometry/uzkujvb6

Thanks!

EDIT: I believe the problem is asking for a straightedge and compass construction.

Answer 1

Hint. Construct any circle tangent to the sides of the angle. Use $\overleftrightarrow{BP}$ to find a point $P'$ on that circle. (There are two choices.) Now, figure out how to construct the "matching" circle through $P$.

Note: Each choice of $P'$ yields a different matching circle.

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Since OP has taken the hint, here's an illustration, with $\bigcirc O$ the "any circle":

Answer 2

Well, without more information about the construction, this is trivially done by drawing two parabolas with a ruler and a string, taking P as the foci, and each AB and BC as directrix.

The two intersections of these parabolas will be the two feasible centers.