Question

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

Answer 1

The equation of the given circle is $2x^2 + 2y^2 - x = 0.$

$2x^2 \+ 2y^2 –x = 0$

$⇒(2x^2 \+ x) + 2y^2 = 0$

$⇒(x^2 – 2 (x) (\dfrac{1}{4}) + (\dfrac{1}{4})^2) + y^2 – (\dfrac{1}{4})2 = 0$

$⇒(x – \dfrac{1}{4})^2 \+ (y – 0)^2 = (\dfrac{1}{4})^2$

The standard equation of the circle is $(x-h)^2 +(y-k)^2 = r^2$

$\therefore$ $h = \dfrac{1}{4}, k = 0,$ and $r = \dfrac{1}{4}$

Thus, the circle's center is ($\dfrac{1}{4}, 0$), while its radius is $\dfrac{1}{4}$. (Ans)