Question

Another medal:

http://gyazo.com/78c4560b519a5440928c2e5ed21743cc

Answer 1

@triciaal

Answer 2

There's to many of this test that I do not know ;c. I'm sorry .

Answer 3

do you know the equation for a circle?
a circle with center (h, k)
(x-h)^2 + (y-k)^2 = r^2 r is the radius

Answer 4

I'm not familiar with it.

Answer 5

subtract a negative will be positive

Answer 6

substitute the numbers for h and k in this formula and read all the post

Answer 7

What do you mean?

Answer 8

openstudy is messing up :/

Answer 9

Well, that's a pretty long introduction, but it should cover it up.

A circle is commonly defined as `all the points that are in the distance 'radius' from the 'center' point`.
If you want to make an equation in a coordinate system then you need to make an equation such that only the coordinates that are in the right distance from the center will fit in the equation and make it true.
Usually people use the cartesian coordinate system (which means the axes are perpendicular).

First, we need an expression for distance between 2 points.
Say we have these 2 points A and B in a cartesian coordinate system:

These points have coordinates in form of an 'ordered pair' of (x,y).
As seen in the picture, we can take the difference of the 2 points' components, and form a right angle triangle with the sides \(\Delta x\) (=delta_x) and \(\Delta y\) (=delta_y).
Basically:
$$
\Delta x = B_x - A_x \\
\Delta_y = B_y - A_y
$$
The distance between the points is the hypotenuse of this right angle triangle, so using the pythagorean theorem we can find it:
$$
\text{distance_between_A_B} = \sqrt{(\Delta x)^2 + (\Delta y)^2}
$$ which can also be written as
$$
(\text{distance_between_A_B})^2 = (\Delta x)^2 + (\Delta y)^2
$$Notice that the order we subtracted the components is not important. if we swap the order then, say, \(\Delta x\) will be negative, but the square of it will remain the same.

We can use this expression to make a 'circle equation'.
We want an equation that is true for all points that are in the distance 'radius' from the 'center' point. Means we have to know both to do so.
Basically, we will take an arbitrary point (x,y), use it along with the center point in the distance equation and set the expected distance to be the radius.
Only the points which are at the right distance from the center will make this equation true.
Let C be the center point, and R the radius of the circle.
$$
\Delta x = x - C_x \\
\Delta y = y - C_y \\
R^2 = (x-C_x)^2 + (y-C_y)^2
$$Which is the circle equation =)
You can compare this to the answers in your question.

~~~ A brief example: ~~~
Say we want an equation for circle with radius of 4 and center at (2,3):
$$
R = 4 \\
C = (2,3) \\
R^2 = (x-C_x)^2 + (y-C_y)^2 \implies 4^2 = (x-2)^2 + (y-3)^2 \\
(x-2)^2 + (y-3)^2 = 16
$$
Now let's make a small test. Let's take the point of the circle which is right above the center.
This point has the same 'x' as the center, so for the distance to be 4 the 'y' should be incremented by 4.
So from the center (2,3) we get to (2,7) which is on the top of the circle. If we plug that into the equation we get:
$$
16 = (2-2)^2 + (3-7)^2 \\
16 = 0^2 + 4^2 \\
16 = 16
$$Which is true statement, so the equation does work for that point.

However, the point (5,2) is not on the circle:
$$
16 = (2-5)^2 + (3-2)^2 \\
16 = (-3)^2 + 1^2 \\
16 = 9 + 1 \\
16 \neq 10
$$And indeed the equation doesn't work for this point. (because the distance from the center is \(\sqrt{10}\) and not 4)

You can see the graph of this example here:
http://www.wolframalpha.com/input/?i=%28x-2%29%5E2+%2B+%28y-3%29%5E2+%3D+16

Hope it helps.

Answer 10

Apparently 'cover up' are not the words I was looking for.. 'It should cover it all' is better. hehe