Question

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

Answer 1

For circle BD is tangent to BQP Secant
$\therefore B{D}^2=BQ.BP=5\times 21=105$
Now draw a tangent from A to circle = AT
$AT=BD(\because$ tangents are drawn to circle from A & B)
Which are equidistant from D, are equal)
Now $A{T}^2=B{D}^2=105$ ....(1)
$A{T}^2=A\times AD$ ....(2)
Consider $\Delta ABD,$
$\frac{AD}{sin\angle OBD}=\frac{AB}{sin\angle ADB}$
$sin\angle OBD=\frac{OD}{OB}=\frac{2}{13}orsin\angle ADB=sin({90}^{\circ }+\angle ODA)=cos\angle ODAcos\angle ODA=\frac{DY}{OD}=\frac{XD}{2\times 8}=\frac{XD}{16}\therefore \frac{AD}{\frac{8}{13}}=\frac{26}{\frac{XD}{16}}OrAD.XD=\frac{\text{/}{26}^2\times 8\times 16}{\text{/}13\times \text{/}16}=256AD(AD-AX)=256A{D}^2-AD.AX=256A{D}^2-105=256orA{D}^2=361\Rightarrow AD=19cm$