Question

AB is the diameter of a circle. P is a point on it such that AP = 24 cm and PB = 6 cm. Find the length of the shortest chord through P.

Answer 1

To find the shortest chord through point P on a circle where AB is the diameter, and AP = 24 cm, PB = 6 cm:

Total diameter AB = AP + PB = 24 cm + 6 cm = 30 cm

So, radius OM = 15 cm.

Point P lies on diameter AB. The shortest chord through P will be the chord perpendicular to AB (diameter) passing through P.

Distance from center O to P = OP = OA − AP = 15 − 24 = −9 cm

(This negative result shows that P is beyond the center, so we can simply take OP = 24 − 15 = 9 cm.)

In right triangle formed with center O and perpendicular chord:

`MP = sqrt(OM^2 - OP^2)`

`sqrt(15^2 - 9^2)`

= `sqrt(225 - 81)`

= `sqrt144`

 = 12 cm