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Question

Two concentric circles are of radii $$5cm$$ and $$3cm$$. The length of the chord of the larger circle which touches the smaller circle  is


A
4cm
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B
12cm
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C
2cm
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D
8cm
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Solution

The correct option is C 8cm
$$ Given-\\ O\quad is\quad the\quad centre\quad of\quad two\quad concentric\quad circles\quad$$ 

$$ { C }_{ 1 },\quad the\quad inner\quad circle\quad with\quad radius=OP=3cm\quad $$and 

$$ { C }_{ 2 },\quad the\quad outer\quad circle\quad with\quad radius=OA=5cm.$$ 

$$AB\quad is\quad a\quad chord\quad of\quad { C }_{ 2 }.\quad It\quad touches\quad { C }_{ 1 }\quad at\quad P.$$ 

$$ To\quad find\quad out-$$ 

$$ The\quad length\quad of\quad AB=?$$  

$$Solution-$$

 $$We\quad know\quad that\quad the\quad radius\quad through\quad the\quad point\quad of\quad$$ $$contact\quad of\quad a\quad tangent\quad to\quad a\quad circle\quad is\quad perpendicular\quad \\ to\quad the\quad tangent.$$ 

$$\therefore \quad OP\bot AB\Longrightarrow \angle OPA={ 90 }^{ o }.........(i)$$ 

$$ Again\quad we\quad know\quad that\quad he\quad perpendicular,\quad dropped\quad from\quad the\quad \\ center\quad of\quad a\quad circle\quad to\quad \quad any\quad of\quad its\quad chord,\quad bisects\quad the\quad latter.$$ 

$$ \therefore \quad OP\quad bisects\quad AB\quad at\quad P.$$ 

$$So\quad AB=2\times AP.......(ii).$$          
 
$$ Now\quad \Delta OAP\quad is\quad a\quad right\quad triangle\quad with\quad OA\quad as\quad hypotenuse$$ 

$$ (from\quad i).$$

$$ OA=5cm\quad \& \quad OP=3cm.$$ 

$$ So,\quad applying\quad Pythagoras\quad theorem,\quad we\quad have$$ 

$$AP=\sqrt { { OA }^{ 2 }-{ OP }^{ 2 } } =\sqrt { { 5 }^{ 2 }-{ 3 }^{ 2 } } cm=4cm.$$ 

$$\therefore \quad AB=2\times AP(from\quad ii)=2\times 4cm=8cm.\\ Ans-\quad Option\quad D.\\ \\ \\ \\ \\  $$

272038_238246_ans.png

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