Question

which equation represents the general form a circle with a center at (-2, -3) and a diameter of 8 units?\
\\( x^2 + y^2 + 4x + 6y - 51 = 0 \\)\
\\( x^2 + y^2 - 4x - 6y - 51 = 0 \\)\
\\( x^2 + y^2 + 4x + 6y - 3 = 0 \\)\
\\( x^2 + y^2 - 4x - 6y - 3 = 0 \\)

Explanation:

Step1: Find radius from diameter

Radius $r = \frac{\text{diameter}}{2} = \frac{8}{2} = 4$

Step2: Write standard circle equation

Standard form: $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)=(-2,-3)$
$\implies (x+2)^2 + (y+3)^2 = 4^2$

Step3: Expand the standard equation

Expand $(x+2)^2$: $x^2 +4x +4$
Expand $(y+3)^2$: $y^2 +6y +9$
Sum: $x^2 +4x +4 + y^2 +6y +9 = 16$

Step4: Simplify to general form

Combine constants: $4+9-16 = -3$
$\implies x^2 + y^2 +4x +6y -3 = 0$

Answer:

$x^2 + y^2 + 4x + 6y - 3 = 0$ (third option)