If it is possible to draw a line which belongs to all the given family of lines `y-2x+1+lambda_1(2y-x-1)=0,3y-x-6+lambda_2(y-3x+6)=0`,`a x+y-2+lambda_3(6x+a y-a)=0` , then
(a)`a=4` (b) `a=3` (c)`a=-2` (d) `a=2`

Length of the shortest chord of the parabola y^2=4x+8 , which belongs to the family of lines (1+lambda)y+(lambda-1)x+2(1-lambda)=0 is

The line common to the given families of line a(x +y+1)+b(2x-3y-8)=0 and lambda(x-y-1)+mu (2x-y-1)=0 is : x+y+1=0 x+y-1=0 x+y-1=0 None of these

The equation of straight line belonging to both the families of lines (x-y+1)+lambda_1(2x-y-2)=0 and (5x+3y-2)+lambda_2(3x-y-4)=0 where lambda_1, lambda_2 are arbitrary numbers is (A) 5x -2y -7=0 (B) 2x+ 5y - 7= 0 (C) 5x + 2y -7 =0 (D) 2x- 5y- 7= 0

The value of the lambda if the lines (2x+3y+4)+lambda(6x-y+12)=0 are

The centre of family of circles cutting the family of circles x^2 + y^2 + 4x(lambda-3/2) + 3y(lambda-4/3)-6(lambda+2)=0 orthogonally, lies on

The equation of the locus of the foot of perpendicular drawn from (5, 6) on the family of lines (x-2)+lambda(y-3)=0 (where lambda in R ) is

Consider the family of lines 5x+3y-2+lambda_(1)(3x-y-4)=0 " and " x-y+1+lambda_(2)(2x-y-2)=0 . Find the equation of a straight line that belongs to both the families.

The number of circles belonging to the system of circles 2(x^(2)+y^(2))+lambda x-(1+lambda^(2))y-10=0 and orthogonal to x^(2)+y^(2)+4x+6y+3=0 , is

The equation of the circle which cuts orthogonally the three circles x^2+y^2+4x+2y+1=0, 2x^2+2y^2+8x+6y-3=0, x^2+y^2+6x-2y-3=0 is

If the lines x - 2y - 6 = 0 , 3x + y - 4=0 and lambda x +4y + lambda^(2) = 0 are concurrent , then value of lambda is