A) rectangle |
B) square |
C) circle |
D) parallelogram |
C) circle |
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.