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

The key idea is to rewrite the given equation into the standard form of a circle: $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the centre and $r$ is the radius.

Step 1: Divide the entire equation by 2 to simplify:

Step 2: Group the $x$ terms and complete the square:

Take half of $-\frac{1}{2}$, square it: $\left(-\frac{1}{4}\right)^2 = \frac{1}{16}$. Add and subtract $\frac{1}{16}$:

$\left(x^2 - \frac{x}{2} + \frac{1}{16}\right) + y^2 = \frac{1}{16}$ …

The equation $2x^2 + 2y^2 - x = 0$ is not in the standard form of a circle. The standard form is $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the centre and $r$ is the radius. To get there, we need to isolate the squared terms and complete the square for the $x$ terms.

Notice that both $x^2$ and $y^2$ have the same coefficient (2). That’s a good sign — it means the equation can represent a circle, not an ellipse. If the coefficients of $x^2$ and $y^2$ were different, we’d be dealing with something else entirely.

Let’s work through it step by step.

1. Divide through by 2 to make the coefficients of $x^2$ and $y^2$ equal to 1.

2. Group the $x$ terms together and prepare to complete the square.

3. Complete the square for $x$. Take half of the coefficient of $x$, which is $-\frac12 \div 2 = -\frac14$, and square it: $\left(-\frac14\right)^2 = \frac{1}{16}$. Add and subtract $\frac{1}{16}$ inside the bracket:

The first three terms form a perfect square:

4. Move the constant to the right side.

Now the equation is in standard form. Compare with $(x-h)^2 + (y-k)^2 = r^2$:

• $h = \frac14$, $k = 0$, so centre is $\left(\frac14, 0\right)$.
• $r^2 = \frac{1}{16}$, so $r = \frac14$ (radius is always positive). …