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A circle having centre as O' and radius ...

A circle having centre as O' and radius r' touches the incircle of `Delta ABC` externally at. F, where F is on BC and also touches its circumcircle internally at G. It O is the circumcentre of `Delta ABC` and I is its incentre, then

A

OO'=R-r'

B

Perpendicular distance from O to line joining IO' is `|(b-c)/(2)|`

C

Projection of OO' on line joining IO'=r'+R cos A

D

`r'=(Delta)/(a)tan^(2)A`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
From figure, OO' = OG - O'G = R-r'

`OE=DE=|BF-BD|=|s-b-(a)/(2)|=(|c-b|)/(2)`
OD = EF = R cos A
From `Delta OEO'`, using Pythagoras theorem, we get
`(R-r')^(2)=(R cos A+r')^(2)+((c-b)/(2))^(2)`
`rArr r' =(bc)/(4R)"tan"^(2)(A)/(2)`
`= (Delta)/(a)"tan"^(2)(A)/(2)`
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