Question

which equation represents the general form a circle with a center at (-2,-3) and a diameter of 8 units?
○ x² + y² + 4x + 6y - 51 = 0
○ x² + y² - 4x - 6y - 51 = 0
○ x² + y² + 4x + 6y - 3 = 0
○ x² + y² - 4x - 6y - 3 = 0

Explanation:

Step1: Write the standard form of the circle equation

The standard form of a circle equation is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center and \(r\) is the radius. Given the center \((-2,-3)\) and diameter \(d = 8\) (so radius \(r=\frac{d}{2}=4\)).
Substitute \(h=-2\), \(k = -3\), and \(r = 4\) into the standard form:

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Step2: Expand the equation

Expand \((x + 2)^2=x^{2}+4x + 4\) and \((y + 3)^{2}=y^{2}+6y+9\).

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Step3: Convert to general form

Subtract \(16\) from both sides: \(x^{2}+y^{2}+4x + 6y+13-16=0\), which simplifies to \(x^{2}+y^{2}+4x + 6y- 3=0\)

Answer:

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