A) (23, 31)
B) (2, 17)
C) [13, 23]
D) [12, 21]
Correct Answer: D
Solution :
Let \[{{S}_{1}}:{{x}^{2}}+{{y}^{2}}-2x-2y+1=0\] \[{{C}_{1}}(1,1),{{r}_{1}}=1\] and \[{{S}_{2}}:{{x}^{2}}+{{y}^{2}}-18x-2y+78=0\] \[{{C}_{2}}(9,1),{{r}_{2}}=2\] Centre of circles lie on opposite side of line \[3x+4y-\lambda =0\] \[\therefore \]\[(3+4-\lambda )(27+4-\lambda )<0\] \[\Rightarrow \]\[(\lambda -7)(\lambda -31)<0\]\[\Rightarrow \]\[\lambda \in (7,31)\] Line lies outside the circles \[{{S}_{1}}\]and \[{{S}_{2}}.\] \[\therefore \]\[\left| \frac{3+4-\lambda }{5} \right|\ge 1\Rightarrow \lambda \in (-\infty ,2]\cup [12,\infty )\] and \[\left| \frac{27+4-\lambda }{5} \right|\ge 2\Rightarrow \lambda \in (-\infty ,21]\cup [41,\infty )\] So,\[\lambda \in [12,21]\]You need to login to perform this action.
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