The equation of a circle is given as 2x\(^2\) + 2y\(^2\) – x – 3y – 41 = 0. Find the coordinates of its centre.
- (\(\frac{-1}{4}\), \(\frac{3}{4}\))
-
(\(\frac{1}{4}\), \(\frac{3}{4}\))
- (\(\frac{-1}{2}\), \(\frac{3}{2}\))
- (\(\frac{-1}{2}\), \(\frac{-3}{2}\))
The correct answer is: B
Explanation
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 → - (\frac{-3}{4}\) = (\frac{3}{4}\)
therefore the centre is (\(\frac{1}{4}\), \(\frac{3}{4}\))