☕ Hello

I was doing an example when this question came to my mind.

The question I was dealing with: two dots (A and B) exist on the paper. Draw a circle with a radius of X which passes through the points (A and B )

The answer: so I know that the center of the circle must be on the perpendicular bisector of line segment A and B

🤔 My question : can the center of the circle be that far which part of the circle becomes exactly on the line And can the center of our circle be that far which it doesn't even pass through the points ?

  • $\begingroup$ The condition on $A,B$ and $X$ for such a circle to exist is if and only if $X\ge\frac12\overline{AB}$. If $X>\frac12\overline{AB}$ there are exactly two such circles, otherwise $1$. $\endgroup$ Oct 27, 2022 at 16:20
  • $\begingroup$ Forgive me what do you mean by there are exactly two such circles,otherwise 1 ? $\endgroup$ Oct 27, 2022 at 16:26
  • $\begingroup$ I mean what it means: if $X>\frac12\overline{AB}$ there are exactly two circles with the property you've said, and if $X=\frac12\overline{AB}$ there is exactly one circle with the property you've said. $\endgroup$ Oct 27, 2022 at 16:42
  • $\begingroup$ If you're speaking of circles in the Euclidean plane, then "no" to both: Each point on the perpendicular bisector is the center of a unique circle through both points, and the union of these circles turns out to be the plane with the (open) segment $\overline{AB}$ removed. $\endgroup$ Oct 27, 2022 at 20:34


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