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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

Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/oo-583684" style="font-weight:bold;" target="_blank" title="Click to know more about OO">OO</a>'=R-r'<br/>Perpendicular distance from O to line joining IO' is `|(b-c)/(2)|`<br/>Projection of OO' on line joining IO'=r'+R cos A <br/>`r'=(<a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a>)/(a)tan^(2)A`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :From <a href="https://interviewquestions.tuteehub.com/tag/figure-987693" style="font-weight:bold;" target="_blank" title="Click to know more about FIGURE">FIGURE</a>, OO' = OG - O'G = R-r' <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CEN_TRI_DPP_6_3_E01_298_S01.png" width="80%"/> <br/> `OE=DE=|BF-BD|=|s-b-(a)/(2)|=(|c-b|)/(2)` <br/> OD = EF = R cos A <br/> From `Delta OEO'`, using Pythagoras theorem, we get <br/> `(R-r')^(2)=(R cos A+r')^(2)+((c-b)/(2))^(2)` <br/> `rArr r' =(<a href="https://interviewquestions.tuteehub.com/tag/bc-389540" style="font-weight:bold;" target="_blank" title="Click to know more about BC">BC</a>)/(4R)"tan"^(2)(A)/(2)` <br/> `= (Delta)/(a)"tan"^(2)(A)/(2)`</body></html>


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