The centre and radius of the circle 2x^2 + 2y | vedclass

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Question

(A) The given equation of the circle is $2x^2 + 2y^2 - x = 0$.Dividing the entire equation by $2$,we get $x^2 + y^2 - \frac{1}{2}x = 0$.Comparing this

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$\left( \frac{1}{4}, 0 \right)$ and $\frac{1}{4}$

Vedclass · Answer

(A) The given equation of the circle is $2x^2 + 2y^2 - x = 0$.Dividing the entire equation by $2$,we get $x^2 + y^2 - \frac{1}{2}x = 0$.Comparing this with the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$,we have $2g = -\frac{1}{2} \implies g = -\frac{1}{4}$,$2f = 0 \implies f = 0$,and $c = 0$.The centre of the circle is $(-g, -f) = \left( \frac{1}{4}, 0 \right)$.The radius $R$ is given by $\sqrt{g^2 + f^2 - c} = \sqrt{\left( -\frac{1}{4} \right)^2 + 0^2 - 0} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.Thus,the centre is $\left( \frac{1}{4}, 0 \right)$ and the radius is $\frac{1}{4}$.

The centre and radius of the circle $2x^2 + 2y^2 - x = 0$ are

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