Yes. The perpendicular bisector of a chord forms a radius when extended to the centre of the circle and a diameter when extended beyond the centre to the opposite point on the circumference.
The perpendicular bisector of ANY chord of the circle goes through the center. Each side of a triangle mentioned would be a chord of the circle therefore it is true that the perpendicular bisectors of each side intersect at the center.
-- Draw any two random chords of the circle. -- Construct the perpendicular bisector of each chord. -- The perpendicular bisectors intersect at the center of the circle. All of this can be done with a compass, an unmarked straight-edge, and a pencil.
yes a diameter passes through the center of a circle
A line that goes through a circle is a secant line. (Remember that a line is infinitely long.) Anyline that passes through a circle is a secant line, whether it passes through the center of the circle or not. Compare this to a line segment with endpoints on the circumference of a circle. That segment is called a cord of that circle. If the cord of a circle passes through the center of that circle, it is a diameter of that circle, which is the longest cord of the circle.
In the case of a circle: Diameter: The distance through the circle - from one end to the other, passing through the center. Circumference: The distance around the circle.
Yes, in a circle, the perpendicular bisector of a chord does indeed pass through the center of the circle. This is because the perpendicular bisector of a chord divides it into two equal segments and is equidistant from the endpoints of the chord. Since the center of the circle is the point that is equidistant from all points on the circle, it must lie on the perpendicular bisector. Thus, any chord's perpendicular bisector will always intersect the center of the circle.
Yes, the perpendicular bisector of a cord is the shortest distance from the centre of a circle to the cord.
A circle itself does not form a perpendicular bisector because a perpendicular bisector is a line that divides a segment into two equal parts at a right angle, typically associated with straight segments. However, the concept of a perpendicular bisector can be applied to chords within a circle. The perpendicular bisector of a chord will always pass through the center of the circle.
The perpendicular bisector of ANY chord of the circle goes through the center. Each side of a triangle mentioned would be a chord of the circle therefore it is true that the perpendicular bisectors of each side intersect at the center.
You have points A, B, and C. Using a compass and straight edge, find a perpendicular bisector of AB (that is, a line that is perpendicular to AB and intersects AB at the midpoint of AB. Next, find a perpendicular bisector of BC. The two lines you found will meet at the center of the circle.
A circle can have perpendicular bisector lines by means of its diameter.
The secant of a circle passes through the center of a circle sometimes
A circle cannot form a perpendicular bisector.
-- Draw any two random chords of the circle. -- Construct the perpendicular bisector of each chord. -- The perpendicular bisectors intersect at the center of the circle. All of this can be done with a compass, an unmarked straight-edge, and a pencil.
A line through a circle that does not go through the center of the circle is a secant line. A line through a circle that does go through the center is still a secant line, by the way. Compare this to a line segment that has its two endpoints on the circumference of the circle. That line segment is a cord of the circle. If that cord of the circle passes through the center of the circle, then the cord is a diameter of that circle.
A circle cannot form a perpendicular bisector.
Draw a line from any part on the outside of a circle to the exact center of the circle. * * * * * That is fine if you know where the center is but not much use if you are just given a circle and do not know where the exact centre is. In this case: Draw a chord - a straight line joining any two points on the circumference of the circle. Then draw the perpendicular bisector of the chord. Draw another chord and its perpendicular bisector. The two perpendicular bisectors will meet at the centre.