In the equation (x+2)^2+(y-3)^2=4, the radius of the circle is

From the equation of a circle, we want to get the radius of the same circle.

We will see that the radius is equal to 2.

First, let's describe the general equation of a circle.

For a circle of radius R centered in the point (h, k) the equation is:

(x - h)^2 + (y - k)^2 = R^2

Here, we got the equation:

(x+2)^2 + (y-3)^2 = 4

We can rewrite it to:

(x- (-2))^2 + (y-3)^2 = 2^2

Then we can see that this is a circle of radius equal to 2 centered at (-2, 3), the graph of the circle can be seen below.

Concluding the radius of the circle is equal to 2.

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center = (-2, 3)

radius = 2

Step-by-step explanation:

(x+2)^2   +  (y-3)^2 =  4

Use the formula:   (x - h)^2   +  (y - k)^2  = r^2

where (h, k) = center,   radius = r

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(x - (-2))^2   +  (y - 3)^2 =  4 =  (2)^2

center = (-2, 3)

radius = 2

see attached

step-by-step explanation:

given the function as;

write the equation replacing f(x) with y

form   and fill a table containing values of y with corresponding values of x

you can take a range for   values of x as -3≤x≤3

solving for x=-3

for x=-2

repeat the same procedure to get the values of y for the remaining ,x=-1 to x=3

plot the values of y against values   of x and obtain the following graph