Question

The centre of the circle 3x² + 3y2 +6x-12y-1=0 is d
a) (1, -3)
b) (-1,3)
c) (-1,2)​

Answer 1

Answer:

Answer is (-1,2)

Step-by-step explanation:

Comparing with the general eqn

ax^+ay^+2gx+2fy+c=0

Here a=3, 2g=6 | g=3 ,2f=-12 | f=-6 ,c=-1

Center = (-g/a,-f/a) =(-3/3,6/3)

                               =(-1,2)

Answer 2

The center of the circle 3x² + 3y2 +6x-12y-1=0 is (-1,3). (Option b)

Given:

3x² + 3y² +6x-12y-1=0

To Find:

Center of the circle

Solution:

Given equation of the circle 3x² + 3y² +6x-12y-1=0

Divide eq by 3.

⇒x² + y² +2x-4y-1/3=0

Now comparing the equation with general eq of circle x² + y² +2gx+2fy+c=0

Center of the circle is (-g,-f)

2g=2 then g= 1

and 2f= -6 then f= -3

Therefore, the center of the circle 3x² + 3y2 +6x-12y-1=0 is (-1,3).

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