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C(1) and C(2) are two concentrate circle...

`C_(1) and C_(2)` are two concentrate circles, the radius of `C_(2)` being twice that of `C_(1)`. From a point P on `C_(2)` tangents PA and PB are drawn to `C_(1)`. Prove that the centroid of the `DeltaPAB` lies on `C_(1)`

Text Solution

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Let r be the radius of `C_(1)` and 2r be the radius of `C_(2)`.Then, `OA=OB=r`
and `OP=2r`
Since PA and PB are tangents to `C_(1)`,
`/_OAP=/_OBP=90^(@)`
Let OP meets `C_(1)` at G.
Let `/_OPA=theta`. Then,
`sin theta=(OA)/(OP)=(r)/(2r)=(1)/(2)`
or `theta =30^(@)`
or `/_ AOP =60^(@)`
In `Delta OAC`,
`cos 60^(@)=(OC)/(OA)`
or `(1)/(2)=(OC)/(r) `or `OC=(r)/(2)`
`:. PC=2r-(r)/(2)-(3r)/(2)`
Also, `CG=r-OC=r-(r)/(2)=(r)/(2)`
Clearly, `CG=(1)/(3)PC`
Therefore, G is the centroid of `Delta ABP`.
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