Why do arc equations for x and y give half circles?

[kishtik]

The first general circle formula is,
$$(x-a)^2+(y-b)^2=r^2$$
Where M(a,b) and r:radius.
I understand this well, but when the subject is arcs...
$$(x-a)^2=r^2-(y-b)^2$$
$$x_\textrm{1,2} =a (+-) \sqrt{r^2-(y-b)^2}$$
My teacher said that equations for x1 and x2 were half circles at right and left. But how?
And also the same fo y,
$$y_\textrm{1,2}=b(+-)\sqrt{r^2-(x-a)^2}$$
were the arcs of top half and bottom of the circle. But why?
Any help is appreciated.

 

[HallsofIvy]

You apparently accept that $$(x-a)^2+(y-b)^2=r^2$$
is the equation of a circle.

To get $$x_\textrm{1,2} =a (+-) \sqrt{r^2-(y-b)^2}$$, you solve for x. Of course, with the square root, you have to take + and - to get both roots.

You know, I hope, that x measures right and left on a graph. The point (4,3) is 4 units to the right of the x-axis. The point (-4,3) is 4 units to the left. When you solve any equation for x, the result is "left" or "right". Taking the positive sign is right, negative, left.

y measures up and down so solving for y does the same thing except up and down instead of right and left.

 

[kishtik]

I knew that basics but couldn't put together. Thanks anyway.