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10-1.Circle and System of Circles
medium

વર્તૂળ $ x^2 + y^2 = r^2$  દ્વારા રેખા  $\frac{x}{a}\,\, + \;\,\frac{y}{b}\,\, = \,\,1$ પરના આંતર છેદથી બનતી જીવાની લંબાઈ....

A

$\frac{{2\,\sqrt {{r^2}\,\left( {{a^2}\, + \;{b^2}} \right)\,\, - \,\,{a^2}\,{b^2}} }}{{{a^2}\,\, + \;\,{b^2}}}$

B

$2\,\,\sqrt {\frac{{{r^2}\,\left( {{a^2}\, + \;{b^2}} \right)\,\, - \,\,{a^2}\,{b^2}}}{{{a^2}\,\, + \;\,{b^2}}}} $

C

$\sqrt {\frac{{{r^2}\,\left( {{a^2}\, + \;{b^2}} \right)\,\, - \,\,{a^2}\,{b^2}}}{{{a^2}\,\, + \;\,{b^2}}}} $

D

એકપણ નહિ

Solution

જીવાની લંબાઈ = $2${(ત્રિજ્યા) $^2$ – (કેન્દ્રથી જીવા પરના લંબની લંબાઈ) $^2$}$^{1/2}$

$ = \,\,\,2\,\,\,{\left\{ {{r^2} – \,\,{{\left( {\frac{{ – 1}}{{\sqrt {(1/{a^2})\,\, + \,\,(1/{b^2})} }}} \right)}^2}} \right\}^{1/2}}\,\, = \,\,2\,\,\sqrt {\frac{{{r^2}({a^2} + {b^2})\,\, – \,\,{a^2}{b^2}}}{{{a^2} + {b^2}}}} $

Standard 11
Mathematics

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