Question

Use Pythagoras theorem and write proof of above theorem "the lengths of tangents drawn from an external point to a circle are equal."​

Answer 1

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Answer 2

$\huge\bold\pink\bigstar$$\huge{\overbrace{\underbrace{\purple{ℍoℓα!!!}}}}$$\huge\bold\pink\bigstar$

$\huge\underline{\underline{\bf\black{Question\::-}}}$

$\sf\red{Use\:Pythagoras\:theorem\:and\:write\:proof\:of }$

$\sf\green{above\:theorem\:"the\:lengths\:of\:tangents}$

$\sf\blue{drawn\:from\:an\:external\:point}$

$\sf\orange{to\:a\:circle\:are\: usually\: equal}$

$\huge\underline{\underline{\bf\black{Solution\::-}}}$

$\huge{\orange{\underline{\red{\mathscr{Given :-}}}}}$

$\sf\pink{Two\:tangents\:PA\:and\:P \:to\:a\:circle\:with}$

$\sf\pink{center\:O,\:from\:an\: exterior\: point\:P.}$

$\huge{\orange{\underline{\red{\mathbb{R.T.P :-}}}}}$$\sf\orange{PA\:=\:PB}$

$\huge{\orange{\underline{\red{\mathscr{Proof :-}}}}}$

$\red{In\:triangle\:OAP;\:angle\:OAP\:= 90°}$

$\sf\green{.°.\:AP^2\:=\:OP^2\:-\:OA^2}$

$\sf\blue{[°.°\: Pythagoras\:theorem]}$

$\implies$$\sf\purple{AP^2\:=\:OP^2\:-\:OB^2}$

$\sf\red{[°.°\:OA\:=\:OB,\:radii\:of\:the\:same\:circle]}$

$\sf\pink{=\:BP^2}$

$\implies$$\sf\green{AP^2\:=\:BP^2}$

$\implies$$\sf\blue{PA\:=\:PB}$

$\huge\underline{\underline{\bf\pink{Hence\:Proved}}}$