If the equation (4a – 3) x2 + ay2 + 6x – 2y + 2 = 0 represents a circle, then its centre is

Given that the equation of the circle is:


(4a - 3)x2 + ay2 + 6x - 2y + 2 = 0 ..... (1)


Comparing with the standard equation of circle:


x2 + y2 + 2ax + 2by + c = 0


We get


4a - 3 = a


4a - a = 3


a = 1


Substituting a value in (1), we get


(4(1) - 3)x2 + 1y2 + 6x - 2y + 2 = 0


x2 + y2 + 6x - 2y + 2 = 0


We know that for a circle x2 + y2 + 2ax + 2by + c = 0


Centre = (- a, - b)


Radius =


Centre(C) =



The correct option is (c)

4