Question

which of the following is the standard form of a circle with center (0, -1) and radius 4?
a. (x)^2 + (y + 1)^2 = 16
b. (x)^2 + (y + 1)^2 = 4
c. (x)^2 + (y - 1)^2 = 4
d. (x)^2 + (y - 1)^2 = 16

Explanation:

Step1: Recall the standard form of a 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 \((0,-1)\), the equation is \((x - 0)^2+(y+1)^2=r^2\), which simplifies to \(x^{2}+(y + 1)^{2}=r^{2}\).

Step2: Analyze each option

• Option a: \(x^{2}+(y + 1)^{2}=16\). Here, \(r^{2}=16\), so \(r = 4\).
• Option b: \(x^{2}+(y + 1)^{2}=4\). Here, \(r^{2}=4\), so \(r=\sqrt{4}=2\).
• Option c: \(x^{2}+(y - 1)^{2}=4\). The center here is \((0,1)\) (not \((0,-1)\)), so this is incorrect.
• Option d: \(x^{2}+(y - 1)^{2}=16\). The center here is \((0,1)\) (not \((0,-1)\)), so this is incorrect.

Answer:

A. \(x^{2}+(y + 1)^{2}=16\)