which equation represents the general form a circle with a center at (-2, -3) and a diameter of 8…

Explanation:

Step1: Recall the standard - form of a circle equation

The standard - form of a circle equation is $(x - a)^2+(y - b)^2=r^2$, where $(a,b)$ is the center of the circle and $r$ is the radius. Given the center $(-2,-3)$ and diameter $d = 8$, then the radius $r=\frac{d}{2}=4$.

Step2: Write the standard - form equation

Substitute $a=-2$, $b = - 3$, and $r = 4$ into the standard - form equation: $(x+2)^2+(y + 3)^2=16$.

Step3: Expand the equation

Expand $(x + 2)^2=x^{2}+4x + 4$ and $(y + 3)^2=y^{2}+6y+9$. So, $x^{2}+4x + 4+y^{2}+6y + 9=16$.

Step4: Convert to general form

Combine like - terms: $x^{2}+y^{2}+4x + 6y+4 + 9-16=0$, which simplifies to $x^{2}+y^{2}+4x + 6y-3=0$.

Answer:

$x^{2}+y^{2}+4x + 6y-3=0$ (the fourth option)