MATH SOLVE

2 months ago

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]

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]