Question

Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.​

Answer 1

Answer:

Solution :

It is given that

PA=5cm,PB=3cm and AB=4cm.

⇒PA^2=(5)2=25.

PB^2+AB^2=(3)2+(4)2=25

∴PA^2=PB^2+AB^2

( Angle opposite to side PA must be of 90∘ (converse of phythagora theorem)

•∠PBA=90∘

It is possible only when B coincides with M I.e centre of smaller circle will lie on M.(the line segment joining the centres of two circles bisect the common chord at right angle)

• PQ=2PB ( perpendicular drawn from the centre to the chord bisects the chord)

=2×3=6cm.

Answer 2

Step-by-step explanation:

In ∆APO

AO^2= 5^2= 25

OP^2= 4^2= 16

AP^2= 3^2= 9

OP^2+ AP^2 = 16+9 =25

now, OP perpendicular to AB

AP = 1/2AB (10.3 theorem)

3 = 1/2 AB

3×2 = AB

AB = 6

The length of the common chord is 6 cm