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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 interval |
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Answer» [13, 23] `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` `rArrlambdain(-oo,21]uu[,oo)""...(v)` From Eqs. (iii), (iv) and (v), we get `lambda in [12, 21]` |
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