Find the centre and radius of the circle 2x^{ | vedclass

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Question

(A) The equation of the given circle is $2x^{2} + 2y^{2} - x = 0$. Divide the entire equation by $2$ to normalize the coefficients of $x^{2}$ and $y^{

Vedclass · Accepted Answer

Centre: $(\frac{1}{4}, 0)$,Radius: $\frac{1}{4}$

Vedclass · Answer

(A) The equation of the given circle is $2x^{2} + 2y^{2} - x = 0$. Divide the entire equation by $2$ to normalize the coefficients of $x^{2}$ and $y^{2}$: $x^{2} + y^{2} - \frac{x}{2} = 0$. Complete the square for the $x$ terms: $(x^{2} - \frac{x}{2} + (\frac{1}{4})^{2}) + y^{2} = (\frac{1}{4})^{2}$. This simplifies to: $(x - \frac{1}{4})^{2} + (y - 0)^{2} = (\frac{1}{4})^{2}$. Comparing this with the standard form $(x - h)^{2} + (y - k)^{2} = r^{2}$,we get the centre $(h, k) = (\frac{1}{4}, 0)$ and the radius $r = \frac{1}{4}$.

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

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