Question

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=0

B.) x^2+y^2-4x-6y-51=0

C.) x^2+y^2+4x+6y-3=0

D.) x^2+y^2-4x-6y-3=0

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

To find the equation, we will form the general equation of a circle and then will find the equation that matches the equation we formed.

General equation of a circle

The general equation of a circle is given as

$(x-h)^2+(y-k)^2 = R^2$

where, the (h, k) are the coordinates of the center of the circle, and R is the radius of the circle.

The correct option is c.

Given to us

• circle with a center at (–2, –3)
• diameter of 8 units

To find

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

Radius of the Circle

Given that the diameter of the circle is 8 units. therefore,

$\rm{ Radius\ of\ the\ circle =\dfrac{Diameter}{2} = \dfrac{8}{2}$

Equation of a circle

The equation of the circle that represents the general form of a circle with a center at (–2, –3) and a radius of 4 units can be found by substituting the values in the equation of a circle:

$(x-h)^2+(y-k)^2 = R^2$

Substituting the values,

$[x-(-2)]^2+[y-(-3)]^2 = 4^2\\\\ (x+2)^2+(y+3)^2 = 16$

$(x+2)^2+(y+3)^2 = 16\\\\ (x^2 + 4 + 4x)+(y^2+9+6y) = 16\\\\ x^2 + 4 + 4x+y^2+9+6y = 16\\\\ x^2 + 4x+y^2+6y = 16-4-9\\\\ x^2 + 4x+y^2+6y = 3\\\\ x^2 + 4x+y^2+6y -3 = 0$

Therefore, the equation which will be in the above form will be the equation we want. As we can see the above equation is similar to   option c.

Verification

To verify that we will plot the graph. The image given below represents the general form of a circle with a center at (–2, –3) and a diameter of 8 units.

Hence, the correct option is c.

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