Two concentric ci

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