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The equation of a circle is given as 2x2 + 2y2 - x...

The equation of a circle is given as 2x2 + 2y2 - x - 3y - 41 = 0. Find the coordinates of its centre.

A.

(14, 34)

B.

(14, 34

C.

(12, 32)

D.

(12, 32)

Correct answer is B

2x2 + 2y2 - x - 3y - 41

standard equation of circle
(x-a)2 + (x-b)2 = r2
General form of equation of a circle.
x2 + y2 + 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.
2x2 + 2y2 - x - 3y - 41

= x2 + y2 + 2gx + 2fy + c
2x2 + 2y2 - x - 3y - 41 = 0
divide through by 2

g = 14 ; 2g = 12

f = 34 ; 2f = 32

a = -g  → - 14 ; = 14

b = -f → - (\frac{-3}{4}\) = (\frac{3}{4}\)

therefore the centre is (14, 34)