if the points ( 0,0 ) , ( 0,1 ) , (1, 0 ) , (2a , 3a ) are concylic then

1   'a' can  attain only rational values

2    'a' is irrational

3     no real value of  'a' exists

4     none of these

It's (a)   ..  :)

 the eqn of circle comes as

(x-1/2)^2 + (y-1/2)^2 = 1/2

so subs value of 4th point 

The centre of the circle is (1/2,1/2) and it's eqn is x²+y²-x-y=0.(2a,3a) lies on it

13a²-5a=0a=0 or a=5/13.So,I think the ans is option1.