Question

Two points a and b are 16 cm apart . A circle with radius 17cm is drawn to pass through these points . Find the distance of ab from the centre of the circle

Answer 1

Answer:

The chord AB is at a distance of 15 cm from centre of the circle.

Step-by-step explanation:

Concept used:

The perpendicular drawn from centre of circle to a chord, bisects the chord.

Pythagors theorem:

In a right angled square on the hypotenuse is equal to sum of the squares on the other two sides.

Given:

AB = 16 cm

OC ⊥ AB

Then AC = BC = 8 cm

In ΔOCB

$OB^2=OC^2+BC^2$

$17^2=OC^2+8^2$

$289=OC^2+64$

$289-64=OC^2$

$OC^2=225$

$OC=\sqrt{225}$

$OC=15\:cm$

Therefore, the chord AB is at a distance of 15 cm from centre of the circle.

Answer 2

Answer:

Step-by-step explanation:

Radius = 17 cm

Distance of ab from center of circle = d cm

it will be perpendicular at ab (at midpoint of ab) let call it M

Center of circle = O

OM = d cm  ( to be find)

OA = OB  = Radius = 17 cm

AM = BM = 16/2 = 8 cm

OM² = OA² - AM²

=> d² = 17² - 8²

=> d² = (25)(9)

=> d² = 5² * 3²

=> d = 5 * 3

=> d = 15 cm

the distance of ab from the centre of the circle = 15 cm