It is given that ,

OA = OB = 10

Diameter = AB = 20

OD = 8

Chord BC of large circle is tangent

to smaller circle.

point of contact is D.

OD is perpendicular bisector of BC

In ∆OBD , we have

<BDO = 90°

OB² = OD² + BD²

10² = 8² + BD²

100 - 64 = BD²

BD² = 36

BD = √36

BD = 6

Therefore ,

BC = 2BD

BC = 2 × 6 = 12

ii ) In ∆ACD ,

<ACB = 90° ( angle in semicircle )

∆ACB right angled triangle.

AB² = AC² + BC²

20² = AC² + 12²

AC² = 20² - 12²

AC² = ( 20 + 12 ) ( 20 - 12 )

= 32 × 8

AC = √ 256

AC = 16

I hope this helps you.

: )