Write the equation in standard form for the circle passing through (0, 23 /2 ) centered at the origin.

Answer 1

The standard form for a circle centered at some point ( h, k ) with a radius of R is:

( x - h )² + ( y - k )² = R²

Since the prompt said "centered at origin", this is math-talk for centered at the center of all graphs: (0, 0)

So, h = 0 and k = 0

So the new standard form equation for this problem's circle is:

x² + y² = R²

For the circle to pass through a point, it MUST continue to satisfy the standard form equation. If you attempt to put in x and y coordinates that do NOT lie on the edge of the circle, you will get the left side of the equation NOT equal to the ride side.

But when points do lie on or pass through the edge of the circle, then they do satisfy the standard form equation.

So, x = 0 and y = 23 / 2

0² + ( 23 / 2 )² = R² R² = 529 / 4 R = ± 23 / 2

Now, I've kept with the convention that when you take a square-root, you do have to consider it could be positive or negative because squaring a negative gets you the positive result again.

But with most circles, negative radius is next to meaningless unless you deal with higher difficulty situations. Thus, you can take the positive result and say:

R = + 23 / 2

And finally, the full standard form of the circle that passes through ( 0 , 23 / 2 ) and is centered at the origin is:

x² + y² = ( 23 / 2 )²

I hope this helps!