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$\lambda$ के सभी वास्तविक मानों का समुच्चय, जिनके लिए वृत्तों $x^{2}+y^{2}-4 x-4 y+6=0$ तथा $x^{2}+y^{2}-10 x-10 y+\lambda=0$ पर ठीक दो उभयनिष्ठ स्पशरेखाएँ खींची जा सकती हों, का जो अंतराल है, वह है
$(12, 32)$
$(18, 42)$
$(12, 24)$
$(18, 48)$
Solution
The equation of the circles are
${x^2} + {y^2} – 10x – 10y + \lambda = 0\,\,\,\,\,\,\,……\left( 1 \right)$
and ${x^2} + {y^2} – 4x – 4y + 6 = 0\,\,\,\,\,\,\,……\left( 2 \right)$
${C_1} = \,$ center of $\left( 1 \right) = \left( {5,5} \right)$
${C_2} = \,$ center of $\,\left( 2 \right) = \left( {2,2} \right)$
$ = {C_1}{C_2} = \sqrt {9 + 9} = \sqrt {18} $
${r_1} = \sqrt {50 – \lambda } ,{r_2} = \sqrt 2 $
For exactly two common tangents we have
${r_1} – {r_2} < {C_1}{C_2} < {r_1} + {r_2}$
$ \Rightarrow \sqrt {50 – \lambda } – \sqrt 2 < 3\sqrt 2 < \sqrt {50 – \lambda } + \sqrt 2 $
$ \Rightarrow \sqrt {50 – \lambda } – \sqrt 2 < 3\sqrt 2 $ or $\,3\sqrt 2 < \sqrt {50 – \lambda } + \sqrt 2 $
$ \Rightarrow \sqrt {50 – \lambda } < 4\sqrt 2 $ or $2\sqrt 2 < \sqrt {50 – \lambda } $
$50 – \lambda < 32$ or $8 > 50 – \lambda $
$ \Rightarrow \lambda < 18$ or $\lambda < 42$
Required interval is $(18,42)$