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10-1.Circle and System of Circles
normal
Circles ${x^2} + {y^2} + 2gx + 2fy = 0$ and ${x^2} + {y^2}$ $ + 2g'x + 2f'y = $ $0$ touch externally, if
A
$f'g = g'f$
B
$fg = f'g'$
C
$f'g' + fg = 0$
D
$f'g + g'f = 0$
Solution
(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}} \times \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'$.
$\therefore {(gf' – fg')^2} = 0$
$\Rightarrow gf' = fg'$.
Standard 11
Mathematics
