Question

what is x^2 + y^2 - 6x - 4y + 4 = 0 in standard form for a circle?

Answer 1

Standard form of the circles equation is:
\[(x-\alpha)^2+(y- \beta)^2=r^2\]
Where alpha and beta are the coordinates of the center and r, is the radius.
You can transform the general equation to the standard form by completing the squares, which is solely based on the form \((x - a)^2=x^2-2ax+a^2\)

Answer 2

Take the x^2 term and the x term and complete the square in x.
Take the y^2 term and the y term and complete the square in y.

Answer 3

how do i do that?

Answer 4

You add what is necessary for any associable group of terms to form a perfect square.

Answer 5

In general, if you have \(x^2 + ax\), and you want to complete the square, you need to add the square of half of the middle coefficient. By middle coefficient I mean \(a\) because it will become the middle coefficient once you add the constant term.
In other words, to complete the square here
\(x^2 + ax\), you add \((\dfrac{a}{2})^2\).
Example:
Complete the square:
\(x^2 + 6x \)
Take half of 6 and square it.
\((\dfrac{6}{2})^2 = 3^2 = 9\)
You get:
\(x^2 + 6x + 9\)

Answer 6

Remember that if you are completing the square in an equation, you need to add the same to both sides of the equation because the rules of equations still apply.

Answer 7

Example:
Complete the square of the equation
\(x^2 + 8x + 7 = 0\)
First, subtract 7 from both sides:
\(x^2 + 8x = -7\)
Now add the correct "complete the square term".
In this case we need \((\dfrac{8}{2})^2 = 4^2 = 16\)

\(x^2 + 8x + 16 = -7 + 16\)

\((x + 4)^2 = 9\)

Answer 8

Do you understand now how completing the square works?

Answer 9

not really

Answer 10

I have broken it down to simple steps.
Tell me which step you do not understand.

Start with

\(x^2 + 8x\)

Take half of 8 which is 4.
Square 4 which is 16.
You need to add 16 to complete the square.

Then you get

\(x^2 + 8x + 16\)