MHT CET202615 April 2026Morning ShiftMathematicsCircleActual
The centre and radius of the circle (a+1)x^2 + 3y^2 - 6x + 9y + a + 4 = 0 are respectively ...
Options
- A(-1, 3 2 ), 5 2
- B(-1, - 3 2 ), 5 2
- C(1, - 3 2 ), 5 2
- D(1, 3 2 ), 5 2
Correct answer
C. (1, - 3 2 ), 5 2
Step-by-step solution
For the given equation to represent a circle, the coefficients of x^2 and y^2 must be equal. a + 1 = 3 a = 2 Substituting a = 2 into the given equation: 3x^2 + 3y^2 - 6x + 9y + 6 = 0 Dividing by 3 , we get the standard form of the circle: x^2 + y^2 - 2x + 3y + 2 = 0 Comparing with x^2 + y^2 + 2gx + 2fy + c = 0 , we have 2g = -2 , 2f = 3 , and c = 2 . The centre of the circle is (-g, -f) = (1, - 3 2 ) . The radius of the circle is r = g^2 + f^2 - c = (-1)^2 + ( 3 2 )^2 - 2 = 1 + 9 4 - 2 = 5 4 = 5 2 . Answer: (1, - 3