Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
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