Compute the centre and radius of the circle 2x2 + 2y2 – x = 0

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Given that, the circle equation is 2x^2+ 2y^2– x = 0

This can be written as:

⇒ (2x^2-x) + y2= 0

⇒ 2{[x^2– (x/2)] +y^2} = 0

⇒{ x^2– 2x(¼) + (¼)^2} +y^2– (¼)^2= 0

Now, simplify the above form, we get

⇒(x- (¼))^2+ (y-0)^2= (¼)^2

The above equation is of the form (x – h)^2+ (y – k)^2= r^2

Therefore, by comparing the general form and the equation obtained, we can say

h= ¼ , k = 0, and r = ¼.

Thanks