In the given figure, CD is the diameter of a circle with centre O and CD is perpendicular to chord AB. If AB = 12 cm and CE = 3 cm, then radius of the circle is
Given: AB = 12cm, CE = 3cm
AB = AE + EB
AE = EB (OC is perpendicular bisector to AB)
∴ AE = 6 cm
Let CD = 2x (diameter)
AO = OC = x (radius)
In Δ AOE
AO2 = AE2 + OE2
x2 = 62 + (OC – EC)2
x2 = 62 + (x – 3)2
x2 = 62 + x2 + 32 – 2(x)(3)
x2 = 36+ x2 + 9 – 6x
6x = 36+ 9 + x2 – x2
6x = 45
x = 
= 7.5
∴ Radius = x = 7.5 cm
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