Question

prove that out of all Chords which passes through any point in circle, that chord will be smallest which is perpendicular on diameter which passes that point

Answer 1

Let C(O, r) is a circle and let M is a point within it where O is the centre and r is the radius of the circle.

Let CD is another chord passes through point M.

We have to prove that AB < CD.

Now join OM and draw OL perpendicular to CD.

In right angle triangle OLM,

OM is the hypotenuse.

So OM > OL

=> chord CD is nearer to O in comparison to AB.

=> CD > AB

=> AB < CD

So all chords of a circle of a circle at a given point within it, the smallest is one which is bisected at that point.