Let f(x,y)=0 be the equation of circle.If `f(0,lambda)=0` has equal roots `lambda=1,1` and `f(lambda,0)=0` has roots `lambda=(1)/(5),5` ,then the radius of the circle

Let f(x,y) =0 be the equation of a circle. If f (0, lamda)=0 has equal roots lamda=1,1 and f (lamda, 0 ) =0 has roots lamda =1/5 ,5, then the radius of the circle is

Let phi (x,y)=0 , be the equation of circle. If phi (0,lambda)=0 has equal roots lambda =2,2 and phi (lambda,0)=0 has roots lambda=4/5, 5 then the centre of the circle is :

Let f(x,y) =0 be the equation of a circle. If f (0, lamda)=0 has equal roots lamda=1,1 and f (lamda, 0 ) =0 has roots lamda =1/2 ,2, then the centre of the circle is

Suppose f(x,y)=0 is the equation of a circle such that f(x,1)=0 has equal roots (each equal to 2) and f(1,x)=0 also has equal roots (each equal to zero). The equation of circle is

f(x, y) = 0 is a circle such that f(0, lambda) = 0 and f(lambda, 0)=0 have equal roots and f(1, 1)= -2 then the radius of the circle is : (A) 4 (B) 8 (C) 2 (D) 1

Let f(x,y)=0 be equation of circle such that (0,0) satisfy it and have family of linces x-2-(lambda+3)y+lambda x as its normal,lambda in R if diameter of circle is k then k^(2)=

If the equation x^(2)+lambda x+mu=0 has equal roots and one root of the equation x^(2)+lambda x-12=0 is 2 ,then (lambda,mu) =

If the equation x^(3)-6x^(2)+9x+lambda=0 has exactly one root in (1, 3), then lambda belongs to the interval

If alpha is a complex number and x^(2)+alpha x+bar(alpha)=0 has a real root lambda then lambda=