The general form of the equation of a circle is x2 + y2 + 8x + 22y + 37 = 0. the equation of this circle in standard form is will medal,

The standard form of the equation for a circle is (x-a)^2 + (y-b)^2 = r^2 where (a,b) = center of circle r = radius. So let's do it. x^2 + y^2 + 8x + 22y + 37 = 0 Let's start with determining a. We know that (x-a)^2 will result in an x^2 term and an 8x term. Let's square (x-a) and we get x^2 - 2ax + a^2. The x^2 term is OK and we have 8x = -2ax So 8x = -2ax 8 = -2a -4 = a So we now know that part of the equation will be (x - -4)^2 = (x + 4)^2. Let's expand that, giving x^2 + 8x + 16 and since we're adding 16 to the left, we need to add that to the right as well. So our equation becomes (x + 4)^2 + y^2 + 22y + 37 = 16 Now let's work on the y terms. Using the same logic as above the equation to solve is 22y = -2by 22 = -2b -11 = b So (y - -11)^2 = (y+11)^2. And expanding gives us y^2 + 22y + 121 which means we need to add 121 to the right, so we have (x + 4)^2 + (y + 11)^2 + 37 = 16 + 121 (x + 4)^2 + (y + 11)^2 + 37 = 137 (x + 4)^2 + (y + 11)^2 = 100 And a simple rewrite of the right hand side gives us (x + 4)^2 + (y + 11)^2 = 10^2 So the final equation is (x + 4)^2 + (y + 11)^2 = 10^2 Which indicates a circle with a radius of 10 centered at (-4,-11)

The answer to the question is d