Question

what is the standard form of the equation of a circle with center (3, -2) and radius 4?

• $(x - 3)^2 + (y - 2)^2 = 16$
• $(x - 3)^2 + (y + 2)^2 = 16$
• $(x - 3)^2 + (y + 2)^2 = 4$
• $(x + 3)^2 + (y - 2)^2 = 16$

Explanation:

Step1: Recall the standard form of a circle's equation

The standard form of the equation of a circle with center \((h, k)\) and radius \(r\) is \((x - h)^2+(y - k)^2=r^2\).

Step2: Identify \(h\), \(k\), and \(r\) from the problem

Given center \((3, - 2)\), so \(h = 3\) and \(k=-2\). The radius \(r = 4\), then \(r^2=4^2 = 16\).

Step3: Substitute \(h\), \(k\), and \(r^2\) into the standard form

Substitute \(h = 3\), \(k=-2\), and \(r^2 = 16\) into \((x - h)^2+(y - k)^2=r^2\). We get \((x - 3)^2+(y-(-2))^2=16\), which simplifies to \((x - 3)^2+(y + 2)^2=16\).

Answer:

B. \((x - 3)^2+(y + 2)^2=16\)