Question

Find the center and radius of the circle whose equation is x 2 + y 2 - 6x - 2y + 4 = 0

Answer 1

So, you would want to put this into the vertex form. That would tell you the information you want by observation.
Do you know how to do that?

Answer 2

Just to comfirm, by x2 you mean \[x ^{2}\]?

If that is what it means then this is what you do. First I would put all the terms with an x and all the terms with a y into little groups. \[(x ^{2} -6x) + (y ^{2} -2y) +4 = 0\]

Then you do something called completing the square for both the brackets. So to complete the square we half the single x and y terms and take the squares out of the bracket.
\[(x -3)^{2} + (y -1) ^{2} +4 =0\]

However when we multiple out those brackets we will get a 9 and a 1 appearing that weren't there before. (-3x-3 =9) (-1x-1=1) So to counteract those terms we add a -9 and a -1 \[(x -3)^{2} + (y -1) ^{2} +4-1-9 =0\]

Finally we just rearrange to get all the numbers on one side\[(x -3)^{2} + (y -1) ^{2} =6\]

This is now in the form of the equation of a circle. So the centre is (3,1) and the radius is root 6.

Answer 3

no i dont

Answer 4

these are the answers i have to choose from.
(3,1), r = sqrt 6
(-3,-1), r = sqrt 6
(3,1) r = 6
(9,1), r = 36

Answer 5

We would need a method called completing the square.

We want to create perfect square trinomials out of the terms with x and the terms with y:
x^2 - 6x, y^2 - 2y
The method is fairly straightforward, we divide the coefficient multiplied to x by 2 and then square the result. That will need to be added and subtracted because we can factor the first part:
\(\color{blue}{x^2 - 2h x + h^2} - h^2 = \color{blue}{(x - h)^2} -h^2 \)

In the problem, we take -6x. Its coefficient is -6. We divide it by two, then square the result:
-6 / 2 = -3; (-3)^2 = 9.
The same process applies to the -2y term.
So if we add and subtract 9 from the x-part, \(\color{blue}{x^2 - 6x + 9} - 9 \), we make a completed square: \( \color{blue}{(x - 3)^2} - 9\)

Answer 6

So to show that in context,
\( x^2 + y^2 - 6x - 2y + 4 = 0 \)
Group the x and y terms:
\( x^2 - 6x + y^2 - 2y + 4 = 0 \)
We add and subtract that which makes a perfect square, which was discussed to be 9.
\( x² - 6x \color{lime}{+ 9 - 9} + y² - 2y + 4 = 0\)
\( \color{blue}{(x^2 - 6x + 9)} - 9 + y^2 - 2y + 4 = 0 \)
\( \color{blue}{(x - 3)^2} - 9 + y^2 - 2y + 4 = 0\)
If you can do this same process for the y quadratic, you will be able to move your constants onto the right and finish!