Question

Which equation represents the general form a circle with a center at (–2, –3) and a diameter of 8 units?

Answer 1

First you need to use the standard form of a circle and then transform it into the general form of a circle.

Standard Form of a Circle.
$(x - h)^2 + (y - k)^2 = r^2$

To use the standard form of a circle, we need to know the center of the circle and the radius of the circle. Our center is at (-2, -3), which is (h, k). Our radius is half of the diameter, which the diameter is 8 so 8 / 2 = 4. So we have a radius of 4. 
In our Standard Form of a Circle, the h = -2 and the k = -3 and the 4 = r
Now we can turn this into the general form of a circle.

Insert our point (-2, -3) and our radius of 4 into the standard form of a circle equation and transform it into the general form.

$(x - h)^2 + (y - k)^2 = r^2$
$(x - (-2))^2 + (y - (-3))^2 =4^2$
$(x + 2)^2 + (y + 3)^2 = 4^2$
$(x + 2)^2 + (y + 3)^2 = 16$
Expand (x + 2)^2
$x^2 + 4x + 4 + (y + 3)^2 = 16$
Expand (y + 3)^2
$x^2 + 4x + 4 + y^2 + 6y + 9 = 16$
Move the 16 over to the left side by adding a - 16
$x^2 + 4x + 4 + y^2 + 6y + 9 - 16 = 16 - 16$
$x^2 + 4x + 4 + y^2 + 6y + 9 - 16 = 0$
Now arrange and combine like terms
$x^2 + 4x + 4 + y^2 + 6y + 9 - 16 = 0$
Gathering all constant values
$x^2 + 4x + y^2 + 6y + 9 -16 + 4 = 0$
Gathering all x and y values
$x^2 + 4x + y^2 + 6y + 9 - 16 + 4 = 0$
$x^2 + y^2 + 4x + 6y + 9 - 16 + 4 = 0$
Combining all constant terms
$x^2 + y^2 + 4x + 6y + 9 - 16 + 4 = 0$
$x^2 + y^2 + 4x + 6y - 3 = 0$
There is nothing else to combine so we have the general form of a circle, which is $x^2 + y^2 + 4x + 6y - 3 = 0$ so the answer is C

Answer: C
$x^2 + y^2 + 4x + 6y - 3 = 0$

Answer 2

The answer is letter C.
Solution:
x^2 + y^2 +4x +6y -3 = 0
x^2 +4x +__ +y^2 +6y +__ = 3 +__ +__
x^2 +4x +4 +y^2 +6y +9 = 3 +4 +9
x^2 +4x +__ +y^2 +6y +__ = 3 +__ +__
(x + 2)^2 + (y + 3)^2 = 16
(x + 2)^2 + (y + 3)^2 = 4^2
(x - h)^2 + (y - k)^2 = r^2
center = -2, -3
2r = diameter
r = 4
2(4) = 8
diameter = 8