**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 + OB^{2} = AO^{2}

⇒ AB^{2} = AO^{2} – OB^{2}

⇒ AB= √(AO^{2} – OB^{2})

⇒ AB= √(5^{2} – 3^{2})

⇒ AB= √(25 – 9)

⇒ AB = √16

⇒ AB = 4 cm

**Similarly,**

By Pythagoras Theorem in ∆COB,

AB^{2} + OB^{2} = CO^{2}

⇒ CB^{2} = CO^{2} – OB^{2}

⇒ CB = √(CO^{2} – OB^{2})

⇒ CB = √(5^{2} – 3^{2})

⇒ CB = √(25 – 9)

⇒ CB = √16

⇒ CB = 4 cm

**Now, **

AC = AB + BC

= 4 cm + 4 cm

= 8 cm

**Hence, Length of chord = 8 cm**