Q:

Which equation represents the general form a circle with a center at (–2, –3) and a diameter of 8 units?A.) x^2+y^2+4x+6y-51=0B.) x^2+y^2-4x-6y-51=0C.) x^2+y^2+4x+6y-3=0D.) x^2+y^2-4x-6y-3=0

Accepted Solution

A:
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.
[tex](x - h)^2 + (y - k)^2 = r^2[/tex]

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.

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

Answer: C
[tex]x^2 + y^2 + 4x + 6y - 3 = 0[/tex]