Question

two circles of radii 10 cm and 8cm intersect and the length of the common cord is 12cm. Find the distance between their centres.

Answer 1

Let O and O' be the centers of the circle of radii 10 cm and 8 cm respectively.

Let PQ be their common chord.

We have,

OP = 10 cm

O'P = 8 cm

PQ = 12 cm

(Perpendicular from the center of the circle to a chord bisects the chord).

In right OLP, we have

OP2 = OL2 + PL2

In right O'LP we have

(O'P)2 = (PL)2 + (O'L)2

OO' = 8 + 5.29

= 13.29 cm

Answer 2

In triangle AOB

AB= 10 radius

BO = 12/2 = 6

As perpendicular drawn from centre bisects the chord

So

AO = √AB^2 - BO^2 = √10^2 - 6^2

= √100 - 36 = √64 = 8

In BOD

BD = 8

BO = 6

OD = √BD^2 - BO^2 = √8^2 -6^2

= √64 -36 = √28 = 2√7

So distance between centres are

AO + OD = AD = 8 +2√7 = 8 +5.29 = 13.29cm