All chords have the same length, so their midpoints are at the same distance from the centre of the circle. Therefore, the midpoints of all such chords form a circle concentric with the given circle.
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In a rectangle, the diagonals are equal and bisect each other. Since the rectangle is cyclic, each diagonal is a diameter of the circle. Therefore, their intersection is the midpoint of a diameter, which is the centre.
Let ABCD be a parallelogram inscribed in a circle. Opposite angles of a cyclic quadrilateral are supplementary, while opposite angles of a parallelogram are equal. Therefore, each angle is 90°. Hence, ABCD is a rectangle.
The perpendicular distances of the two chords from the centre are found using the Pythagoras Theorem. Since the chords lie on opposite sides of the centre, the required distance is the sum of these distances, which is seven centimetres.
In an isosceles triangle, the altitude from the vertex bisects the base. The perpendicular bisector of a chord passes through the centre of the circle. Therefore, the altitude passes through the centre.