The answer will depend on the location of the points B, P and C.
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It is a RayA ray is part of a line which is finite in one direction, but infinite in the other. It can be defined by two points, the initial point, A, and one other, B. The ray is all the points in the line segment between A and B together with all points, C, on the line through A and B such that the points appear on the line in the order A, B, C.Source: Faber, Richard L. (1983). Foundations of Euclidean and Non-Euclidean Geometry. New York, United States: Marcel Dekker. ISBN 0-8247-1748-1.
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An arc.
exactly three times as far from point A as they are from point B?
distance
The analogy for d-B to r is like comparing the distance between two points on a straight line (d-B) to the radius of a circle (r). Just as the radius measures the distance from the center of a circle to any point on its circumference, d-B represents the shortest distance between two points on a line.
The answer will depend on the location of the points B, P and C.
It takes 3 non collinear points to define one specific circle. With only two points an infinite number of circles can be drawn. Proof: Given two points A, B draw the line between them. Then find the perpendicular bisector of the line AB. Any point on the perpendicular bisector is equidistant from the two original points, A and B. A circle with center C and radius AC will then pass through points A and B. There are infinite point C's on the perpendicular bisector so there are infinite circles. Given three points A, B and D you can find the perpendicular bisector for line segements AB and then the perpendicular bisector fof line segment BC. The two perpedicular bisectors will not be parallel because the points A, B and D are non collinear. This means the two perpeniducar bisectors will intercept at only one point C(like any two intercepting lines). This point C is equidistant from points A, B, and D. A circle with center C and radius AC will then pass through all three of the points. Since there is only one point C that lies on both perpendicular bisectors, there is only one circle possible.
a circle 9 cm from point b I was co fused by this but you just do a diagram and write this
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This starts with the collocation circle to go through the three points on the curve. First write the equation of a circle. Then write three equations that force the collocation circle to go through the three points on the curve. Last, solve the equations for a, b, and r.
It is a RayA ray is part of a line which is finite in one direction, but infinite in the other. It can be defined by two points, the initial point, A, and one other, B. The ray is all the points in the line segment between A and B together with all points, C, on the line through A and B such that the points appear on the line in the order A, B, C.Source: Faber, Richard L. (1983). Foundations of Euclidean and Non-Euclidean Geometry. New York, United States: Marcel Dekker. ISBN 0-8247-1748-1.
There cannot be such a postulate because it is not true. Consider a line segment AB and let C be any point on the line between A and B. If the three points are A, B and C, there can be no circle that goes through them. It is so easy to show that the postulate is false that no mathematician would want his (they were mostly male) name associated with such nonsense. Well, if one of the points approach the line that goes through the other two points, the radius of the circle diverges. The line is the limit of the ever-growing circles. In the ordinary plane, the limit itself does not exist as a circle, but mathematicians have supplemented the plane with infinity to "hold" the centres of such "straight" circles.