Sure! Let's go through the steps to find the equation of a circle in its general form, given that the circle is centered at [tex]\((-2, -3)\)[/tex] and has a diameter of 8 units.
### Step 1: Identify the Radius
The diameter of the circle is 8 units. The radius [tex]\( r \)[/tex] is half of the diameter:
[tex]\[ r = \frac{8}{2} = 4 \][/tex]
### Step 2: Write the Standard Form of the Circle's Equation
The standard form of the equation of a circle with center [tex]\((h, k)\)[/tex] and radius [tex]\( r \)[/tex] is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
Plug in the values for the center [tex]\((h, k) = (-2, -3)\)[/tex] and the radius [tex]\( r = 4 \)[/tex]:
[tex]\[ (x - (-2))^2 + (y - (-3))^2 = 4^2 \][/tex]
[tex]\[ (x + 2)^2 + (y + 3)^2 = 16 \][/tex]
### Step 3: Expand the Equation
Expand the left-hand side of the equation:
[tex]\[ (x + 2)^2 + (y + 3)^2 = 16 \][/tex]
[tex]\[ x^2 + 4x + 4 + y^2 + 6y + 9 = 16 \][/tex]
### Step 4: Simplify the Expanded Equation
Combine all the constant terms and simplify:
[tex]\[ x^2 + y^2 + 4x + 6y + 4 + 9 = 16 \][/tex]
[tex]\[ x^2 + y^2 + 4x + 6y + 13 = 16 \][/tex]
### Step 5: Rewrite in General Form [tex]\(Ax^2 + By^2 + Cx + Dy + E = 0\)[/tex]
Move all the terms to one side of the equation to get it into the general form:
[tex]\[ x^2 + y^2 + 4x + 6y + 13 - 16 = 0 \][/tex]
[tex]\[ x^2 + y^2 + 4x + 6y - 3 = 0 \][/tex]
Thus, the equation in the general form that represents the circle is:
[tex]\[ x^2 + y^2 + 4x + 6y - 3 = 0 \][/tex]
So, the correct answer is:
[tex]\[ \boxed{x^2 + y^2 + 4x + 6y - 3 = 0} \][/tex]
