Question

Easy question: How do you make this equation equal to y, as in the equation of a line? x=sqrroot (4-y^2)

Answer 1

how did the y loose its square?

Answer 2

\[x=\sqrt{4-y^2}\]
\[x^2=4-y^2\]
\[x^2+y^2=4\]

Answer 3

it is the right half of a circle with center and the origin and radius 2

Answer 4

if you want to solve for y be careful
start with
\[x^2+y^2=4\]
\[y^2=4-x^2\]
\[y=\pm\sqrt{4-x^2}\]

Answer 5

thanks! just one question though. By right half do you mean quadrants 1 and 4?

Answer 6

the equation for the circle with center (0,0) and radius 2 is
\[x^2+y^2=4\] if you solve this for x you actually get
\[x=\pm\sqrt{4-y^2}\]

Answer 7

note the important "plus or minus" because of course you do not know which

Answer 8

if you just write
\[x=\sqrt{4-y^2}\]without the plus or minus, that means you are forcing x to be positive, because that is what the radical sign means. so if x is positive it is the right half of the circle. quadrants one and 4 yes.

Answer 9

if you write
\[y=\sqrt{4-x^2}\] you get the upper half of the circle

Answer 10

so should I graph the upper or right side?