Question

Given two concentric circles of radii 5 and 3, find the length
of a chord of larger circle which touches the smaller one.
If BD = 5 find BC.​

Answer 1

Given:

AO (say) = CO (say) = 5 cm

BO (say) = 3 cm

Let AC be the tangent which meets the circle at the point B and O be the center of circle.

Property: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

By above property, ∆AOB is right-angled at ∠OBA and ∆COB is right-angled at ∠OBC.

Therefore,

By Pythagoras Theorem in ∆AOB,

AB2 + OB2 =AO2

⇒ AB 2 = AO2 – OB 2

⇒ AB= √(AO2 – OB 2)

⇒ AB= √(52 – 32)

⇒ AB= √(25 – 9)

⇒ AB = √16

⇒ AB= 4 cm

Similarly,

By Pythagoras Theorem in ∆COB,

AB2 + OB2 =CO2

⇒ CB 2 = CO2 – OB 2

⇒ CB= √(CO2 – OB 2)

⇒ CB= √(52 – 32)

⇒ CB= √(25 – 9)

⇒ CB = √16

⇒ CB= 4 cm

Now,

AC = AB + BC

= 4 cm + 4 cm

= 8 cm

Hence, Length of chord = 8 cm

Answer 2

Answer:

Given:

R of larger circle = 5cm

R of smaller circle = 4cm

Construction:

Draw OM perpendicular to AB touching AB at M

Now AM=BM

Proof:

In figure we have right triangle OBM

By pythagoras theorem,

BM=3cm

We know that perpendicular drawn from centre to a chord bisects the chord. It means that M is the midpoint of chord AB

Therefore AB= 2AM

AB=2×3

AB=6cm

HOPE IT HELPS..