# Find the centre and radius of the circle $2x^2+2y^2-x = 0$

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• Given C (h, k) be the centre and r the radius of circle. Let P(x, y) be any point on the circle. Then, by the definition, | CP | = r . By the distance formula, we have, $(x-h)^2+(y-k)^2=r^2$
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Given: the equation of the circle, we need to rewrite it in the form $(x-h)^2+(y-k)^2=r^2$, so that we can determine the centre as (h,k) and radius as $r$
Given $2x^2+2y^2-x = 0$
$\quad \quad 2(x^2-\large\frac{1}{2}$$x) + 2y^2 = 0 \rightarrow (x^2-\large\frac{1}{2}$$x) + 2y^2 = 0$
$\quad \quad (x^2-2x\large\frac{1}{4}$$+ \large\frac{1}{4}^2) + y^2 - \large\frac{1}{4}^2 = 0 \quad \quad (x-\large\frac{1}{4} ) ^2$$ + (y-0)^2 = \large\frac{1}{4}^2$
$\Rightarrow$ centre = $(\large\frac{1}{4}$$,\;0)$ and radius = $\large\frac{1}{4}$
answered Apr 1, 2014
edited Apr 1, 2014