Math, asked by shashanknagar3329, 10 months ago

The center of the circle 3x^2+3y^2+6x-12y-6=0 is what

Answers

Answered by rishu6845

Answer:

(-1 , 2 )

Step-by-step explanation:

Given---> Equation of circle

3x² + 3y² + 6x - 12y - 6 = 0

To find---> Coordinate of center of circle

Solution---> Standard Equation of circle

x² + y² + 2gx + 2fy + C = 0

Then coordinate of center of circle = (-g , -f )

So we first divide given equation by 3 to change it in to standard form and then compare it with standard equation to find coordinate of center of circle

3x² + 3y² + 6x - 12y - 6 = 0

Dividing whole equation by 3 , we get

3x²/3 + 3y²/3 + 6x/3 - 12y/3 - 6/3 = 0

=> x² + y² + 2x - 4y -2 = 0

Now comparing it with x² + y² + 2gx + 2fy + C =0

2g = 2 , 2f = -4 , C= -2

g = 2 / 2 f = -4 / 2

g = 1 f = -2

Coordinate of center of the circle = ( -g , -f )

= ( - 1 , 2 )

Answered by Aɾꜱɦ

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(-1 2)

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