The middle points of all chords (each having the same length) of a circle lie on a

A) rectangle

B) square

C) circle

D) parallelogram

Correct answer:
C) circle

Description for Correct answer:

Let 0 \( \Large \left(\alpha ,\ \beta\right) \) and r be the centre and radius of the given circle respectively.



Let M (h. k) be the middle point of the chord AB of length 2d Clearly, \( \Large \left(2d < 2r, or \  d < r\right) \).

Since, M is the middle point of AB, therefore OM is perpendicular to AB.

Therefore, \( \Large OA^{2}=AM^{2}+OM^{2} \)

=> \( \Large r^{2}=d^{2}+ \left(h-a\right)^{2}+ \left(k - \beta \right)^{2} \)

Hence M lies on the circle,

\( \Large \left(x- \alpha \right)^{2}+ \left(y- \beta \right)^{2} = r^{2} - d^{2} \)

It has the same centre but less radius than the given circle.



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