Name the points as shown

Draw \(OH\perp BC\),then H is the midpoint of BC.So BH = 1 cm

Quadrilateral ABHK has 3 right angles at A,B and H

=> ABHK is a rectangle => HK = AB = 2 cm

Let r be the radius of the circle,so OH = KH - OK = 2 - r (cm)

\(\Delta OHB\) right at H has : OH^{2} + HB^{2} = OB^{2} (Pythagoras theorem)

\(\Rightarrow\left(2-r\right)^2+1^2=r^2\Rightarrow4-4r+r^2+1-r^2=0\)

\(\Rightarrow5-4r=0\Rightarrow r=\dfrac{5}{4}\)

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