In the adjoining figure, BD is the diameter of the circle which bisects the chord AC at point E. If `AC=8cm`, `BE=2cm`, then find the radius of the circle.

Explanation

Let radius `OA=OB =r`
`OE=OB-EB`
`=(r-2)cm`
`AE=(AC)/(2)=(8)/(2)=4cm`
Now, `angle OEA=90^@`
(`because` The line segment joining the centre to mid-point of a chord is perpendicular to the chord)
`therefore` In `DeltaOAE`,
`OA^2=OE^2+AE^2` (by Pythagoras theorem)
`rArrr^2=(r-2)^2+4^2`
`rArrr^2=r^2-4r+4+16`
`rArr4r=20 cm`
`rArr=5cm`
`therefore` Radius of circle =5 cm .