Constructing a Circle Given Another Circle and Three Points

Question

Given a circle $o$ and distinct points $A, B,S$ outside $o$, how can I construct a circle $o'$ through $A$ and $B$, such that $S$ lies on the line joining the two intersection points of $o$ and $o'$

Any hints,

Thanks

Answer 1

Well, such circle $\omicron'$ might not exist. However, there is a solution if you extend the problem in the following form: Given a circle $\omicron$ and distinct points $A, B$ and $S$ outside $\omicron$, construct a circle $\omicron'$ through $A$ and $B$, such that $S$ lies on the radical axis of the two circles $\omicron$ and $\omicron'$.

1. Construct the circle $k_S$ centered at $S$ and orthogonal to $\omicron$.

2. Construct the inverse image $A'$ of $A$ with respect to $k_S$, or alternatively construct the inverse image $B'$ of $B$ with respect to $k_S$ (or you can do both it is still going to work).

3. Draw the unique circle $\omicron'$ passing through $A, B, A'$ (or through $A, B, B'$ which is going to be the same circle $\omicron'$). In any case $\omicron'$ passes through the four points $A, A', B, B'$.

4. If $\omicron$ and $\omicron'$ intersect at two points, then $S$ will lie on the line determined by the two intersection points of $\omicron$ and $\omicron'$. If $\omicron$ and $\omicron'$ touch at one point, then $S$ will lie on the line tangent to both $\omicron$ and $\omicron'$ at their point of contact. If $\omicron$ and $\omicron'$ do not intersect, then $S$ will lie on their radical axis.

In any case, $S$ will lie on the radical axis of $\omicron$ and $\omicron'$.