Circles {x^2} + {y^2} + 2gx + 2fy = 0 and {x^ | vedclass

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Question

(a) According to the figure, $OA + O'A = OO'$

$\sqrt {{g^2} + {f^2}} + \sqrt {f{'^2} + g{'^2}} $

$= \sqrt {{{(g' - g)}^2} +

Vedclass · Accepted Answer

$f'g = g'f$

Vedclass · Answer

(a) According to the figure, $OA + O'A = OO'$

$\sqrt {{g^2} + {f^2}} + \sqrt {f{'^2} + g{'^2}} $

$= \sqrt {{{(g' - g)}^2} + {{(f' - f)}^2}} $

$ \Rightarrow {g^2} + {f^2} + f{'^2} + g{'^2} + 2\sqrt {{g^2} + {f^2}} 	imes \sqrt {f{'^2} + g{'^2}} $

$ = {(g' - g)^2} + {(f' - f)^2}$

$ \Rightarrow 2\sqrt {{g^2} + {f^2}} \sqrt {f{'^2} + g{'^2}} = - 2(gg' + ff')$

$ \Rightarrow {g^2}f{'^2} + {f^2}g{'^2} = 2gg'ff'$.

$	herefore {(gf' - fg')^2} = 0$

$\Rightarrow gf' = fg'$.

Circles ${x^2} + {y^2} + 2gx + 2fy = 0$ and ${x^2} + {y^2}$ $ + 2g'x + 2f'y = $ $0$ touch externally, if

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