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)