CAT 2020 Quant Question [Slot 2] with Solution 23

Given, \(d + 2 = D \Rightarrow r + 1 = R\)

In the figure \(OT = r\) and \(OB = r + 1\)

\(OT \bot AB\) as AB is the tangent

OT bisects AB i.e., \(TB = \frac{6}{2} = 3\)

Now, in \(\Delta OTB,O{T^2} + T{B^2} = O{B^2}\)

\(\therefore {r^2} + {3^2} = {(r + 1)^2} \Rightarrow r = 4\)

\(\therefore D = 2(R) = 2(r + 1) = 10cm\)