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$x^{2}+ y^{2}+ c^{2} =2ax$ અને $x^{2} + y^{2} + c^{2} - 2by = 0$ સમીકરણવાળા વર્તૂળો એકબીજાને બહારથી ક્યારે સ્પર્શેં ?
$\frac{1}{{{b^2}}}\,\, + \,\,\frac{1}{{{c^2}}}\,\, = \,\,\frac{1}{{{a^2}}}$
$\frac{1}{{{c^2}}}\,\, + \,\,\frac{1}{{{a^2}}}\,\, = \,\,\frac{1}{{{b^2}}}$
$\frac{1}{{{a^2}}}\,\, + \;\,\frac{1}{{{b^2}}} = \,\,\frac{1}{{{c^2}}}$
એકપણ નહિ
Solution
$S_1\equiv x^{2} + y^{2} – 2ax + c^{2}= 0;$
$ S_2 \equiv x^{2} + y^{2} – 2by + c^{2} = 0$
${c_1}\, \equiv \,\,(a\,,\,\,0)\,,\,\,\,\,{r_1}\, = \,\sqrt {{a^2} – {c^2}} \,\,;\,\,\,{c_2}\, \equiv \,\,(0\,,\,\,b),\,\,\,{r_2} = \,\,\sqrt {{b^2} – {c^2}} $
જો વર્તૂળો એકબીજાને બહારથી સ્પર્શેં $c_1c_2 = r_1 + r_2$ તો
$\begin{array}{l} \Rightarrow \,\,\sqrt {{a^2} + {b^2}} \,\, = \,\,\sqrt {{a^2} – {c^2}} \, + \,\sqrt {{b^2} – {c^2}} \\ \Rightarrow \,\,{a^2} + {b^2} = {a^2} – {c^2} + {b^2} – {c^2} + 2\sqrt {({a^2} – {c^2})\,\,({b^2} – {c^2})} \\ \Rightarrow \,\,\,{c^2} = \,\,\sqrt {({a^2} – {c^2})\,\,({b^2} – {c^2})} \,\,\,\, \Rightarrow \,\,{c^4} = \,\,{a^2}{b^2} – \,({a^2} + {b^2})\,\,{c^2} + {c^4}\\ \Rightarrow \,\,({a^2} + {b^2})\,{c^2} = \,\,{a^2}{b^2}\,\,\, \Rightarrow \,\,\frac{1}{{{a^2}}}\, + \,\,\frac{1}{{{b^2}}}\,\, + \,\,\frac{1}{{{c^2}}}\end{array}$
