Find the center of the circle given by the equation $x^{2} + y^{2} - 4 x + 6 y - 51 = 0$ .

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Solution

Given equation is
$x^{2} + y^{2} - 4 x + 6 y - 51 = 0$
$\Rightarrow x^{2} + y^{2} - 4 x + 6 y - (64 + 9 - 4) = 0$
$\Rightarrow (x^{2} - 4 x + 4) + (y^{2} + 6 y + 9) - 64 = 0$
$\Rightarrow (x - 2)^{2} + (y + 3)^{2} = 64$

On comparing the given equation with the standard equation of circle i.e., $(x - h)^{2} + (y - k)^{2} = r^{2}$ we get, h = 2 and k = -3.

So, the centre of the circle is (2,-3).

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