# 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

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