Didn't you post this earlier? Saw my explanation?

yepp i saw the one on equation of circle formula but i could not find the explanation on completing the square

The whole idea is to rewrite the equation in a form such that you get to the standard form of the circle.

LockB, it’s more of a “double completing the square”. One for x and one for y.

btw how did they do completing the square on the equation x^2 + y^2..... tho, it kind of looks confusing as it isnt the typical completing the square done on quadratic formula

but the x and y is already combined together tho

Let's first do an example of getting the general form of the equation from the standard form.

Let's say you have a circle with centre (5,3) and radius of 4 units

Now standard form of the equation is
(x - a)² + (y - b)² = r²

With the above info,

(x - 5)² + (y - 3)² = 4²
x² - 2(5)x + 5² + y² - 2(3)y + 3² = 16
x² - 10x + 25 + y² - 6y + 9 = 16
x² - 10x + y² - 6y = 16 - 25 - 9
x² - 10x + y² - 6y = -18

We put all the x together first and all the y together next. We put all constants to the right hand side first.

Then, we do completing the square. I demo you one on this chat next.

Suppose the general equation of a circle is

x² + y² + 2x - 4y - 4 = 0.

Then, we collect all the x first followed by all the y on the left hand side. We bring the constant to the other side.

x² + 2x + y² - 4y = 4

Next, we do a completing the square on the x² + 2x and we do a similar idea on the y² - 4y. But we introduce these same terms on the other side of the equation for equality purposes.

x² + 2x + 1² + y² - 4y + 2² = 4 + 1² + 2²
x² + 2x + 1² + y² - 4y + 2² = 9
(x + 1)² + (y - 2)² = 9
(x + 1)² + (y - 2)² = 3²

where the centre of the circle is (-1, 2) and the radius of the circle is 3.

For the question's equation 2x² + 2y² - 3x + 4y = -1,

we are simply completing the square to get back to the standard form.

What was shown in the working was completing the square for the variable x and y simultaneously.

2x² + 2y² - 3x + 4y = -1

Divide both sides by 2,


x² + y² - 3/2 x + 2y = -½
x² - 3/2 x + y² + 2y = -½

Perhaps you are not used to having simultaneous completing of the square so I'll show you the one for x first.


x² - 2(¾)x + (¾)² - (¾)² + y² + 2y = -½

(x - ¾)² - 9/16 + y² + 2y = -½

Notice I left the terms in y untouched.


Next, complete the square for variable y

(x - ¾)² - 9/16 + y² + 2(1)y + 1² - 1² = -½

(x - ¾)² - 9/16 + (y + 1)² - 1 = -½


Now I bring the remaining constants over to the right side.

(x - ¾)² + (y + 1)² = -½ + 1 + 9/16
(x - ¾)² + (y + 1)² = -8/16 + 16/16 + 9/16
(x - ¾)² + (y + 1)² = 17/16

Finally,


(x - 3/4)² + (y + 1)² = (√(17/16))²

so we can now say :

Centre is (¾ , -1) and radius is √(17/16) = √17 / √16
= √17 / 4

I have shown you a variant of completing the square whereby the terms are rewritten as x² - 2cx + c², which can then be rewritten as (x - c)²


(Recall that :

(a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b² )


Likewise for y , it was rewritten in the form
y² + 2dy + d² , which is then rewritten as (y + d)²

thx for explaining to me :)