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दो वृत्त, जो $(0,a)$ व $(0, - a)$ से गुजरते हैं एवं रेखा $y = mx + c$ को स्पर्श करते हैं, एक-दूसरे को समकोण पर काटेंगे यदि
${a^2} = {c^2}(2m + 1)$
${a^2} = {c^2}(2 + {m^2})$
${c^2} = {a^2}(2 + {m^2})$
${c^2} = {a^2}(2m + 1)$
Solution
(c) वृत्तों का समीकरण है, $({x^2} + (y – a)(y + a)) + \lambda x = 0$
$ \Rightarrow {x^2} + {y^2} + \lambda x – {a^2} = 0$
और $\sqrt {{{\left( {\frac{\lambda }{2}} \right)}^2} + {a^2}} $
$= \frac{{\frac{{ – m\lambda }}{2} + c}}{{\sqrt {1 + {m^2}} }}$
$ \Rightarrow (1 + {m^2}){\rm{ }}\left[ {\frac{{{\lambda ^2}}}{4} + {a^2}} \right] = {\left( {\frac{{m\lambda }}{2} – c} \right)^2}$
$ \Rightarrow (1 + {m^2}){\rm{ }}\left[ {\frac{{{\lambda ^2}}}{4} + {a^2}} \right] $
$= \frac{{{m^2}{\lambda ^2}}}{4} – mc\lambda + {c^2}$
$ \Rightarrow {\lambda ^2} + 4mc\lambda + 4{a^2}(1 + {m^2}) – 4{c^2} = 0$
$\therefore $ ${\lambda _1}{\lambda _2} = 4[{a^2}(1 + {m^2}) – {c^2}] $
$\Rightarrow {g_1}{g_2} = [{a^2}(1 + {m^2}) – {c^2}]$
और ${g_1}{g_2} + {f_1}{f_2} = \frac{{{c_1} + {c_2}}}{2}$
$\Rightarrow {a^2}(1 + {m^2}) – {c^2} = – {a^2}$
${c^2} = {a^2}(2 + {m^2})$.
