The radii of two concentric circles are 13 cm and 8 cm respectively. AB is a diameter of the bigger circle. BD is

Question

The radii of two concentric circles are 13 cm and 8 cm respectively. AB is the diameter of the bigger circle. BD is a tangent to the smaller circle touching it at D. Find the length of AD.

Answer 1

Let the line BD intersect the bigger circle at C. Join AC. Then, in the smaller circle.

OD ⊥ BD (radius ⊥ tangent at the point of contact) 

⇒ OD ⊥ BC ⇒ BD = DC (BC is the chord of the bigger circle and perpendicular from the center of the circle to a chord bisects the chord) 

⇒ D is the mid-point of BC 

Also, given O is the mid-point of AB   (AB is the diameter of the bigger circle) 

∴ In ∆ BAC, O is the mid-point of AB and D is the mid-point of BC.

∴ OD = \(\frac{1}{2}\) AC (segment joining the mid-points of any two sides of a triangle is half the third side)

⇒ AC = 2 OD ⇒ AC = 2 × 8 = 16 cm

In right ∆ OBD OD² + BD² = OB² 

⇒ BD = \(\sqrt{OB^2 - OD^2}\)  = \(\sqrt{(13)^2 -8^2}\) = \(\sqrt{169 -64}\) = \(\sqrt{105}\)  

∴ DC = BD = \(\sqrt{105}\)

Now AD² = AC² + DC² 

⇒ AD² = 16² + (\(\sqrt{105}\))² 

= 256 + 105 

AD² = 361 

⇒ AD = \(\sqrt{361}\)  

= 19 cm