Construct the great circle (geodesic) in spherical or Riemannian geometry

Question

Given:

• a circle $C$ with centre $M$,
• two points $P_1$ and $P_2$ inside circle $C$, so that $M$ is not on the line $P_1P_2$,

construct another circle $O$ so that:

• $P_1$ and $P_2$ are on circle $O$,
• circle $O$ cuts the circle $C$ in the points $Q_1$ and $Q_2$,
• and $M$ (the centre of circle $C$ ) is on the the segment $Q_1Q_2$.

This is a construction in a model of spherical or riemannian geometry (the circle $O$ is the great circle or geodesic through points $P_1$ and $P_2$).

But how does the construction work?

Answer 1

Try it:

1. Draw line $e$ such that $\{P_1,P_2\} \subset e$.

2. Draw line $a$, the perpendicular bisector of $P_1P_2$.

3. Choose a point $E$ on $a$ and draw a circle $d$ with radius $P_1E$ and center $E$.

4. Let $F$ and $G$ the intersection points of $c$ and $d$, draw a line $b$ such that $\{F,G\} \subset b$.

5. Let $\{H\}=b \cap e$, draw a line $f$ through $H$ and $M$.

6. Let $\{Q_1,Q_2\} = f \cap c$, draw the perpendicular bisector $g$ of $Q_1Q_2$.

7. Let $\{I\} = g \cap a$, draw circle $o$ with radius $Q_1I$ and center $I$.

Answer 2

In this diagram, circle $c$ has center $M$, and points $P_1$ and $P_2$ not collinear with $M$ are inside $c$. These objects are colored blue. (My drawing program, Geogebra, uses a different naming convention than you, so I must use slightly different names for my objects than you used. This diagram is more busy than I like: sorry!)

Draw the perpendicular bisector $a$ of line segment $\overline{P_1P_2}$ and mark two arbitrary points on it, $A_1$ and $A_2$. (These points just need to be far enough away from $P_1$ and $P_2$ for the next step to be possible.) For each $A_j$, draw the circle with center $A_j$ through $P_1$ (or $P_2$) and let the intersection points of this circle with circle $c$ be called $R_j$ and $S_j$. Construct the midpoint of segment $\overline{R_jS_j}$ and call it $M_j$.

The key to this construction is that the locus of all such points $M_j$ is a circular arc that goes through the point $M$. Therefore we draw the perpendicular bisectors of line segments $\overline{MM_1}$ and $\overline{MM_2}$ and call the intersection point $U$. Draw circle $k$ (in green) with center $U$ through point $M$ (or $M_1$ or $M_2$, it doesn't matter). The part of this circle that is inside circle $c$ is our desired locus of all points $M_j$.

Draw a line through points $M$ and $U$ and let the intersections of that line with circle $c$ be called $Q_1$ and $Q_2$. Draw the perpendicular bisector of line segment $\overline{Q_2P_2}$ (or of any $P_j$ with any $Q_j$ or even of $Q_1$ with $Q_2$) and let the intersection point of that line with line $a$ be called $A$, colored red.

$A$ is then the center of our desired circle. Draw the circle with center $A$ through any of the points $P_1,P_2,Q_1,Q_2$ and call it $o$, in red. This is the desired circle.

Answer 3

The trick is that the centres of all circles that cut circle $C$ diametrically and go trough Point $P_1$ are on a line ( the line $d$ below )

Draw the ray $a $ from $P_1$ trough $M$

Draw line $b$ trough $M$ perpendicular to $a$

Point $R$ is one of the points where $b$ cuts Circle $C$

Line $c$ is the perpendicular bisector of line segment $P_1R$

Point $T$ is where $c$ cuts ray $a$

Draw line $d$ trough $T$ perpendicular to $a$

Line $e$ is the perpendicular bisector of line segment $P_1P_1$

The intersection of line $d$ and line $e$ is the centre of circle $O$.

it seems to work (but I don't know why)