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Find the centre of the circle

x^2 + y^2 + 4x - 6y = 12

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centre = (- 2, 3 )

Step-by-step explanation:

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k ) are the coordinates of the centre and r is the radius

given

x² + y² + 4x - 6y = 12 ( collect x/ y terms )

x² + 4x + y² - 6y = 12

using the method of completing the square

add ( half the coefficient of the x/ y terms)² to both sides

x² + 2(2)x + 4 + y² + 2(- 3)y + 9 = 12 + 4 + 9

(x + 2)² + (y - 3)² = 25 ← in standard form

with centre (- 2, 3 ) and r = $$\sqrt{25}$$ = 5
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