Question

question 37 of 39
what is the equation of a circle with center (-2, 3) and radius 4?
a. $(x - 2)^2+(y + 3)^2 = 16$
b. $(x + 2)^2+(y - 3)^2 = 4$
c. $(x + 2)^2+(y - 3)^2 = 16$
d. $(x + 2)^2-(y - 3)^2 = 16$

Explanation:

Step1: Recall the standard equation of a circle

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

Step2: Identify \(h\), \(k\), and \(r\)

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

Step3: Substitute values into the standard equation

Substitute \(h=-2\), \(k = 3\), \(r = 4\) into \((x - h)^2+(y - k)^2=r^2\).
We get \((x-(-2))^2+(y - 3)^2=4^2\), which simplifies to \((x + 2)^2+(y - 3)^2=16\).

Answer:

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