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If a variable line,  \mathrm{3 x+4 y-\lambda=0}  is such that the two circles   \mathrm{x^2+y^2-2 x-2 y+1=0} and \mathrm{x^2+y^2-18 x-2 y+78=0} are on its opposite sides, then the set of all values of \mathrm{\lambda}  is in the interval

Option: 1

(23,31)


Option: 2

(2,17)


Option: 3

[13,23]


Option: 4

[12,21]


Answers (1)

best_answer

Let   \mathrm{S_1: x^2+y^2-2 x-2 y+1=0}
\mathrm{ C_1(1,1), r_1=1 \text { and } S_2: x^2+y^2-18 x-2 y+78=0 }
\mathrm{ C_2(9,1), r_2=2 }
Centre of circles lie on opposite sides of line

\mathrm{ 3 x+4 y-\lambda=0 \therefore(3+4-\lambda)(27+4-\lambda)<0 \\ }
\mathrm{ \Rightarrow(\lambda-7)(\lambda-31)<0 \Rightarrow \lambda \in(7,31) }

Line lies outside the circles   \mathrm{ S_1~ and ~S_2 }

\mathrm{ \therefore\left|\frac{3+4-\lambda}{5}\right| \geq 1 \Rightarrow \lambda \in(-\infty, 2] \cup[12, \infty) }

and   \mathrm{ \left|\frac{27+4-\lambda}{5}\right| \geq 2 \Rightarrow \lambda \in(-\infty, 21] \cup[41, \infty) }
So, \mathrm{ \lambda \in[12,21] }

Posted by

HARSH KANKARIA

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