Two concentric circle of radii A …
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Sia ? 6 years, 5 months ago
Let O be the common centre of the two circles
and AB be the chord of the larger circle which touches the smaller circle at C.
Join OA and OC.
Then, OA = a and OC = b.
Now, {tex} \mathrm { OC } \perp A B{/tex} and OC bisects AB [{tex}\because{/tex} the chord of the larger circle touching the smaller circle, is bisected at the point of contact].
In right {tex}\triangle A C O{/tex}, we have
OA2 = OC2+AC2 [by Pythagoras' theorem]
{tex}\Rightarrow{/tex}AC = {tex}\sqrt { O A ^ { 2 } - O C ^ { 2 } } = \sqrt { a ^ { 2 } - b ^ { 2 } }{/tex}.
{tex}\therefore{/tex} AB = 2AC = {tex}2 \sqrt { a ^ { 2 } - b ^ { 2 } }{/tex} {tex}[ \because \text { C is the midpoint of } A B ]{/tex}
i.e, required length of the chord AB = {tex}2 \sqrt { a ^ { 2 } - b ^ { 2 } }{/tex}
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