1.

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

Answer»

[13, 23]
[2,17]
[12,21]
[23,31]

Solution :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`
`rArrlambdain(-oo,21]uu[,oo)""...(v)`
From Eqs. (iii), (iv) and (v), we get
`lambda in [12, 21]`


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