$2x^2 + 2y^2 – x = 0$, find the centre and the radius of the circles.

$\begin {array} {1 1} (A)\;\bigg( -\large\frac{1}{4}, 0\bigg) \: and \: \large\frac{1}{4} & \quad (B)\; \bigg(0, -\large\frac{1}{4}\bigg) \: and \: \large\frac{1}{4} \\ (C)\;\bigg( \large\frac{1}{4}, 0\bigg) \: and \: \large\frac{1}{4} & \quad (D)\;\bigg(0, +\large\frac{1}{4}\bigg) \: and \: \large\frac{1}{4} \end {array}$

Toolbox:
• Equation of a circle with centre (h,k) and radius r is given as $(x-h)^2+(y-k)^2=r^2$
The equation of the given circle is
$2x^2+2y^2-x=0$
Dividing throughout by 2 we get,
$x^2+y^2-\large\frac{x}{2}$$=0 This can be written as \bigg( x^2 -\large\frac{x}{2} \bigg)$$+y^2=0$
(i.e) $\bigg[ \bigg( x - \large\frac{1}{4} \bigg)^2 $$- \large\frac{1}{16} \bigg]$$+ (y-0)^2=0$
(i.e) $\bigg( x - \large\frac{1}{4} \bigg)^2$$+(y-0)^2=\large\frac{1}{16} This is of the form (x-h)^2+ (y-k)^2=r^2 Comparing both the equations, we get, h = \large\frac{1}{4}$$, k = 0$ and $r = \large\frac{1}{4}$
Hence the coordinates of the centre are $\bigg( \large\frac{1}{4}$$, 0 \bigg)$ and radiys $r = \large\frac{1}{4}$