Mht CetCircleMHT CET 2026 15th April Morning ShiftMCQ+2 / -02026The centre and radius of the circle $(a+1)x^2 + 3y^2 - 6x + 9y + a + 4 = 0$ are respectively ...A$\left(-1, \dfrac{3}{2}\right), \dfrac{\sqrt{5}}{2}$B$\left(-1, -\dfrac{3}{2}\right), \dfrac{\sqrt{5}}{2}$C$\left(1, -\dfrac{3}{2}\right), \dfrac{\sqrt{5}}{2}$D$\left(1, \dfrac{3}{2}\right), \dfrac{\sqrt{5}}{2}$Check AnswerClear SelectionReveal AnswerShow Explanation
