Find a point that best fits criteria

kevvvli123 shared this question 1 year ago
Answered

For example, if I have a circle with diameter AB and I want a point P such that AP/BP = 2.

I know this particular example could be done with some geometric properties, but I want a generalized way to do this. Like find a point on a parabola such that the ratio of distance from circle w_1 and w_2 is 11/3 or something.

I just want GeoGebra to test like a million points and find the best one. Is there a way to do this?

Comments (9)

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a point P such that AP/BP = 2

I think there are infinite such points. Do you want them all? (ie a locus)

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Or do you mean it has to lie on the circle?

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AP= ratio*BP is a circle. create it and intersect with the object you want

for another objects think that ie: distance(P,circle)=distance(P,centre)-radius etc

share news construction if you need help

Files: foro.ggb
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sorry


type eq1: (x - x(A))² + (y - y(A))² = ratio^2 * ((x - x(B))² + (y - y(B))²) in foro.ggb attached in previous post for better use of ratio


¡marditas prisas!

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This is actually Apollonius circle construction controlled (by slider-ratio)..GIven segment AB, divide it internally and externally at points C and D using the ratio value and then draw circle with CD as the diameter. Any point P on this circle divides AB in the given ratio.

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.....(continued)... Intersection points with another given curve with Apollonius circle will be the required points in general.

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Yo propuse este método porque se puede adaptar a otras situaciones

por ejemplo a una circunferencia y una recta

Files: foro.ggb
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Thank you Sir, for taking note of me - a novice on this GeoGebra platform. I always keenly study your intelligent solutions.

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