The equation of a circle is given as 2x\(^2\) + 2y\(^2\) - x - 3y - 41 = 0. Find the coordinates of its centre.

  • A (\(\frac{-1}{4}\), \(\frac{3}{4}\))
  • B (\(\frac{1}{4}\), \(\frac{3}{4}\))
  • C (\(\frac{-1}{2}\), \(\frac{3}{2}\))
  • D (\(\frac{-1}{2}\), \(\frac{-3}{2}\))

The correct answer is B. (\(\frac{1}{4}\), \(\frac{3}{4}\))

2x\(^2\) + 2y\(^2\) - x - 3y - 41

standard equation of circle

(x-a)\(^2\) + (x-b)\(^2\) = r\(^2\)

General form of equation of a circle.

x\(^2\) + y\(^2\) + 2gx + 2fy + c = 0

a = -g, b = -f., r2 = g2 + f2 - c

the centre of the circle is (a,b)

comparing the equation with the general form of equation of circle.

2x\(^2\) + 2y\(^2\) - x - 3y - 41

= x\(^2\) + y\(^2\) + 2gx + 2fy + c

2x\(^2\) + 2y\(^2\) - x - 3y - 41 = 0

divide through by 2

g = \(\frac{-1}{4}\) ; 2g = \(\frac{-1}{2}\)

f = \(\frac{-3}{4}\) ; 2f = \(\frac{-3}{2}\)

a = -g  → - \(\frac{-1}{4}\) ; = \(\frac{1}{4}\)

b = -f â†’ - (\frac{-3}{4}\) = (\frac{3}{4}\)

therefore the centre is (\(\frac{1}{4}\), \(\frac{3}{4}\))

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