Question number 8

Ans 0,2,1/3

.................................................

2y= x+3 has intercepts (0, 3/2) and ( – 3, 0).

note that y= kx – 1 has intercept (0, – 1) on y axis.

now, the eqn of circle passing through (0, – 1), (0, 3/2) and ( – 3, 0) can be easily written as

 (x + 1.25)²  +  (y - 0.25)²  =  3.125

now when y=0, x= – 1.25 ± 1.75= – 3, ½ 

but, x intercept of y= kx – 1 is (1/k, 0). so if 1/k= ½ then k= 2. 1/k= – 3 will not be considered as this is already an intecept og 2x= y+3 and in the above calculations we have assumed 4 distinct points.

now, we know that a circle always passes thru 3 non collinear pts. so if x intercept 1/k= – 3 then k= – 1/3 also works.

if k= 0, then there is no x intercept and hence only 3 points.