The graph of a circle describes y as a function of x. 
True 
False 

This is false because in order to have a function, it
must not be possible to draw a vertical line and have
it touch or cross the graph more than just ONE point.

Take this graph of a circle:



Now let's arbitrarily draw a blue vertical line down 
through the graph:



Notice that it intersects the circle at two points one 
above the x-axis and one below the x-axis.  This is not
allowed for the graph of a function, so a circle does not
represent a function, i.e., does not describe y as a
function.

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To answer the next questions, you must memorize that
the standard equation of the circle with center (h,k) 
and radius r is

(x - h)² + (y - k)² = r²

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Which of these equations describes a circle
with radius 5 and center (3,-2) ?
(x - 3)² + (y + 2)² = 25 
(x - 3)² + (y + 2)² = 5 
(x + 3)² + (y - 2)² = 25 
(x + 3)² - (y + 2)² = 25

Substitute h=3, k=-2, r=5 into

(x - h)² + (y - k)² = r²
(x - 3)² + (y -(-2) )² = 5²
(x - 3)² + (y + 2)² = 25
So it's the first choics.

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Which of these equations describes a circle
with radius 3/2 and center (0,4) ?
x² - (y - 4)² = 9/4 
x² + (y - 4)² = 3/2 
(x - 4)² + y² = 9/4 
x² + (y - 4)² = 9/4

Substitute h=0 k=4 r=3/2 into

(x - h)² + (y - k)² = r²
(x - 0)² + (y - 4)² = (3/2)²
      x² + (y - 4)² = 9/5
So it's the fourth choice.

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The circle whose equation is 
x² + 2x + y² + 6y + 6 = 0

has center (1,3) and radius 2 
has center (-1, -3) and radius 2 
has center (-1, -3) and radius 4 
has center (1, 3) and radius 4

Now you must learn to change from standard form
of a circle (where there are no parentheses) to 
the standard form (x - h)² + (y - k)² = r². We
do this by completing the square, very similar 
to the way you completed the square back when
you were solving quadratic equations:

    x² + 2x + y² + 6y + 6 = 0

We get the constant term, 6, off the left side by
subtrracting 6 from both sides:

x² + 2x + 1 + y² + 6y + 9 = -6 + 1 + 9

The x terms and the y terms are together, so we
skip a space after the 2x and after the 6y

x² + 2x     + y² + 6y     = -6       

Now we multiply the coefficient of x, which is 2, by 1/2
That gives us 1.
Now we square that result.
1² = 1.  So we add +1 to both sides of the equation. On
the left we put it in the first space we skipped. We add
it on the end of the right side;

x² + 2x + 1 + y² + 6y     = -6 + 1    

Now we multiply the coefficient of y which is 6 by 1/2
That gives us 3.
Now we square that result.
3² = 9  So we add +9 to both sides of the equation. On
the left we put it in the second space we skipped. We add
it on the end of the right side;

x² + 2x + 1 + y² + 6y + 9 = -6 + 1 + 9

Now we factor the first three terms x² + 2x + 1 = 
(x + 1)(x + 1) = (x + 1)²

Now we factor the last three terms y² + 6x + 9 = 
(y + 3)(y + 3) = (y + 3)²

Now we combine the numbers on the right
 -6 + 1 + 9 = 4

So we now have

(x + 1)² + (y + 3)² = 4

Now we compare this with the standard
form:

(x - h)² + (y - k)² = r²

We see that "+ 1" is in the place of "- h", so
- h = + 1 and therefore h = - 1

We see that "+ 3" is in the place of "- k", so
- k = + 3 and therefore h = - 3

So the center is (h, k) = (-1, -3)

We see that "4" is in the place of "r²", so
r² = 4, and therefore r = 2, or its radius is 2

So the answer is the third choice
"has center (-1, -3) and radius 2" 

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The circle whose equation is x² - 6x + y² + 2y = -4

You forgot to list the choices but you can find out 
what its center and radius are.  Follow the steps 
above and you'll get

(x - 3)² + (y + 1)² = 6
                                            _
and its center is (3,-1) and its radius is Ö6

Edwin