Question 145089
1) The answer is NO! You can check this by using the "vertical line test" in which you draw a vertical line through the circle on the graph containing the circle. 
If the vertical line passes through the graph (circle) only once, then the graph is a function. In the case of a circle, the vertical line would pass through the graph (circle) twice, so the equation represented by the graph is not a function.
2) Find the center and the radius of the circle represented by the equation:
{{{x^2+y^2+10x-4y-7 = 0}}}
The goal is to transform this equation into the standard form of the equation for a circle with center at (h, k) and radius r, {{{(x-h)^2+(y-k)^2 = r^2}}}.  First, add 7 to both sides of the equation.
{{{x^2+y^2+10x-4y = 7}}} Now group the x-terms together and the y-terms together.
{{{(x^2+10x)+(y^2-4y) = 7}}} The next step is to "complete the square" in both the x-terms and the y-terms.  You do this by adding the square of half the x-coefficient: ({{{(10/2)^2 = 25}}}) and likewise for the y-coefficient ({{{(-4/2)^2 = 4}}}) to both sides of the equation.
{{{(x^2+10x+25)+(y^2-4y+4) = 7+25+4}}} Now factor the x- and y-groups and simplify.
{{{(x+5)^2+(y-2)^2 = 36}}} Compare this with the standard form {{{(x-h)^2+(y-k)^2 = r^2}}}
You can see that h = -5, k = 2, and {{{r^2 = 36}}}
So, the center is at (h,k) = (-5,2) and the radius, {{{r = sqrt(36)}}} = 6.