InterviewSolution
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InterviewSolution
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If a variable line, `3x + 4y -lambda = 0` is such that the two circles `x^(2)+y^(2)-2x-2y+1=0` and `x^(2)+y^(2)-18x-2y+78=0` are not its opposite sides, then the set of all values of `lambda` is the intervalA. [13, 23]B. [2,17]C. [12,21]D. [23,31] |
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Answer» Correct Answer - C The given circle, `x^(2)+y^(2)-2x-2y+1=0" "...(i)` and `x^(2)+y^(2)-18x-2y+78=0," "...(ii)` are on the opposite side of the variable line `3x+4y-lambda = 0`. Their centres also lie in the opposite sides of the varible line. `rArr[3(1)+4(1)-lambda][3(9)+4(1)-lambda]lt0` [`therefore ` the points ` P(x_(1), y_(1)) and Q (x_(2), y_(2))` lie on the opposite sides of the line ax + by + c =0, `if (ax_(1)+by_(1)+c)(ax_(2)+by_(2)+c)lt0]` `rArr (lambda-7)(lambda-31)lt0` `rArr lambdain(7, 31)" "...(iii)` Also, we have `|(3(1)+4(1)-lambda)/(5)|gesqrt(1+1-1)` `(therefore" Distance of centre from the given line is greater than the radius, i.e., " (ax_(1)+by_(1)+c)/(sqrt(a^(2)+b^(2)))ge r)` `rArr |7-lambda| ge5rArrlambdain(-oo,2] uu [12,oo)" "(iv)` and `|(3(9)+4(1)-lambda)/(5)|gesqrt((81+1-78))` `rArr|lambda-31|ge10` `rArr lambdain(-oo,21]uu[,oo)" "...(v)` From Eqs. (iii), (iv) and (v), we get `lambda in [12, 21]` |
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