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
B
12cm
C
2cm
D
8cm

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 $${ C }_{ 2 },\quad the\quad outer\quad circle\quad with\quad radius=OA=5cm.$$ $$AB\quad is\quad a\quad chord\quad of\quad { C }_{ 2 }.\quad It\quad touches\quad { C }_{ 1 }\quad at\quad P.$$ $$To\quad find\quad out-$$ $$The\quad length\quad of\quad AB=?$$  $$Solution-$$ $$We\quad know\quad that\quad the\quad radius\quad through\quad the\quad point\quad of\quad$$ $$contact\quad of\quad a\quad tangent\quad to\quad a\quad circle\quad is\quad perpendicular\quad \\ to\quad the\quad tangent.$$ $$\therefore \quad OP\bot AB\Longrightarrow \angle OPA={ 90 }^{ o }.........(i)$$ $$Again\quad we\quad know\quad that\quad he\quad perpendicular,\quad dropped\quad from\quad the\quad \\ center\quad of\quad a\quad circle\quad to\quad \quad any\quad of\quad its\quad chord,\quad bisects\quad the\quad latter.$$ $$\therefore \quad OP\quad bisects\quad AB\quad at\quad P.$$ $$So\quad AB=2\times AP.......(ii).$$           $$Now\quad \Delta OAP\quad is\quad a\quad right\quad triangle\quad with\quad OA\quad as\quad hypotenuse$$ $$(from\quad i).$$$$OA=5cm\quad \& \quad OP=3cm.$$ $$So,\quad applying\quad Pythagoras\quad theorem,\quad we\quad have$$ $$AP=\sqrt { { OA }^{ 2 }-{ OP }^{ 2 } } =\sqrt { { 5 }^{ 2 }-{ 3 }^{ 2 } } cm=4cm.$$ $$\therefore \quad AB=2\times AP(from\quad ii)=2\times 4cm=8cm.\\ Ans-\quad Option\quad D.\\ \\ \\ \\ \\$$Maths

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