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?

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).
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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 $45^\circ$.
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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 $Q$.
