Question 94328
A. Complete the square to find the centre and the radius.
{{{x^2+y^2+6x-4y-3 = 0}}} Add 3 to both sides.
{{{x^2+y^2+6x-4y = 3}}} Group the like-terms together on the left side.
{{{(x^2+6x) + (y^2-4y) = 3}}} Now complete the squares in the x- and y-terms. To do this, you add the square of half the x-coefficient and the square of half the y-coefficient to both sides of the equation. 
For the x's this would be {{{(6/2)^2 = 9}}} and for the y's this would be {{{(-4/2)^2 = 4}}}, so we get:
{{{(x^2+6x+9) + (y^2-4y+4) = 3+9+4}}} Factor the trinomials and simplify.
{{{(x+3)^2 + (y-2)^2 = 16}}} Now compare this with the standard form for a circle with centre at (h, k) and radius r.
{{{(x-h)^2 + (y-k)^2 = r^2}}} and you can see that the centre (h, k) is at (-3, 2) while the radius is {{{sqrt(16)}}} = 4.

Follow this approach on the second problem and see how you do.
If you still have problems, please re-post.