Question

The centre of the circle 3x^2+3y^2-12x+15y-23=0​

Answer 1

Answer:

(2,-(5/2))

Step-by-step explanation:

According to property of circle

The coefficients of x^2 and y^2 must be 1

So, the equation becomes

x^2+y^2-4x+5y-(23/3)=0

(we divided equation by 3)

Comparing it with

x^2+y^2+2gx+2fy+c=0

(General equation of circle)

We get g=-2,f=(5/2)

Hece center for required circle is

(-g, - f) =>(2,-(5/2))

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Answer 2

Answer:

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