construct inverse point with respect to the circle by the use of the compass alone If the given point P lies inside a circle C ，with center O，the circle of radius OP about P intersects C in two points. 
How to construct point P' inverse to point P with respect to the circle C 
by the use of the compass alone?

This question is raised from page 144，What is mathematics—2nd Edition
The following content is stated there:

“If the given point P lies inside C the same construction and proof hold, provided that the circle of radius OP about P intersects C in two points.”

I couldn’t understand why does the same construction and proof hold, could you explain it specifically to me?
 A: According to the Mohr-Mascheroni Theorem and my construction below, this can be done. However, as with most "compass-only" constructions,  the process would probably be extremely convoluted.
Since the proof I have uses some ideas from inversive-geometry, I will post it even though it uses both straight-edge and compass.

The main idea of the construction is that the inverse of a point with respect to a line is just the reflection of the point across the line. First, add $Q$ at $\overrightarrow{OP}\cap C$ (below, on the left), and then consider the inverse with respect to the gray circle centered at $Q$ (below, on the right).
$\hspace{5mm}$
Since $C$ passes vertically through $Q$ on the left, it becomes a vertical line on the right. Since the line that contains $P$ and $Q$ passes horizontally through $Q$ on the left, it becomes a line passing horizontally through $P$ and the image of $\infty$ on the right.
The inverse of $P$ with respect to $C$ on the right is $I$, the reflection of $P$ across $C$. Notice that the blue circle on the right passes through $P$ and intersects both $C$ and the line containing $P$ and the image of $\infty$ at right angles. Since inversion is conformal, the blue circle on the left passes through $P$ and intersects both $C$ and the line containing $P$ and $Q$ at right angles.
Consider the green line on the right. It passes through $P$ where it crosses the line containing $P$ and the image of $\infty$ at $45^\circ$. It intersects $C$ at $R$, where $C$ intersects the blue circle. This means that below, it becomes a green circle, passing through $P$ and $Q$ and crossing the line containing $P$ and $Q$ at $45^\circ$.
$\hspace{3cm}$
Since the green circle passes through $P$ and $Q$ and crosses the line containing $P$ and $Q$ at $45^\circ$, the center of the green circle, $E$, forms the $45{-}45{-}90$ $\triangle PQE$. That is, $E$ lies at the intersection of the red circle whose diameter is $\overline{PQ}$ and the perpendicular bisector of $\overline{PQ}$.
Summary:
Given $O$, $P$, and $C$, extend $\overrightarrow{OP}$ to where it intersects $C$ at $Q$. Draw the perpendicular bisector of $\overline{PQ}$ with $D$ at the midpoint of $\overline{PQ}$. Draw the red circle centered at $D$ and passing through $P$.
Place $E$ at either intersection of the perpendicular bisector of $\overline{PQ}$ and the red circle. Draw the green circle centered at $E$ and passing through $P$. Place $R$ at other intersection of $C$ and the green circle.
Place $F$ at the intersection of the perpendicular bisector of $\overline{PR}$ and the line containing $P$ and $Q$. Draw the blue circle centered at $F$ and passing through $P$. $I$ is at the other intersection of the blue circle and the line containing $P$ and $Q$.
A: The construction in the special case you ask about is very simple:
Draw the circle around $P$ through $O$ and by the given hypothesis it intersects the circle $c$ in two points $X$ and $Y$. Draw circles with centers $X$ and $Y$ through $O$. Call $P'$ the second point of intersection of those circles. Then $P'$ is the inverse of $P$ with respect to the circle $c$:


The reason why this works is as follows:

The point $P'$ lies on the line through $OP$ which is fixed under inversion with respect to $c$. Note that $OP$ is the perpendicular bisector of the segment $XY$.
Since $P$ is the circumcenter of the triangle $OXY$, it is the intersection of the perpendicular bisector $\color{red}{x'}$ of $OX$ and the perpendicular bisector $OP'$ of $XY$. After inversion with respect to $c$ those perpendicular bisectors are the circle $\color{red}{x}$ and the line $OP$, and they intersect at $P'$. Now finish off by symmetry.

Alternatively, as robjohn pointed out, you can use similar triangles, not using inversion at all, apart from the definition: write
$$ 
\frac{|OX|/2}{|OP|} = \frac{|OP'|/2}{|OX|}
$$
in order to see that $P$ and $P'$ are inverse to each other with respect to $c$.
A: Generally to invert a point with respect to a circle, construction steps depend on the position of the point with respect to the circle. If the point lies outside the circle then we follow few steps which can't be followed for the point which lies inside forest.
There exists another construction procedure named Dutta's Construction which resolves this inconvenience, i.e. we can follow same steps to invert a point whether it lies outside or inside the forest, a case distinction as in usual treatments is not needed.
Source Links:
https://forumgeom.fau.edu/FG2014volume14/FG201422.pdf
https://en.wikipedia.org/wiki/Inversive_geometry#Dutta's_construction
