If the equation of a circle is `lambdax^2+(2lambda-3)y^2-4x+6y-1=0,` then the coordinates of centre are

Explanation

To find the coordinates of the center of the circle given by the equation:

\[ \lambda x^2 + (2\lambda - 3)y^2 - 4x + 6y - 1 = 0, \]

we will follow these steps:

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