How do you graph $(x+4)^2+(y-1)^2=9$ ?

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Answer 1 · 2018-07-07T14:41:24

A circle with a radius of $3$ , and its center located at $(-4,1)$ .

Given: $(x+4)^2+(y-1)^2=9$ .

Notice that the equation for a circle is given by:

$(x-a)^2+(y-b)^2=r^2$

where:

$(a,b)$ are the coordinates of the circle's center

$r$ is the radius of the circle

Here, we get $(a,b)=(-4,1)$ , and $9=3^2$ .

So, this equation shows us a circle with a radius of $3$ and has a center located at $(-4,1)$ .

Here is a graph of the circle:

graph{(x+4)^2+(y-1)^2=9 [-10, 10, -5, 5]}

Answer 2 · 2018-07-07T15:19:44

See below:

The good thing is that this equation is in standard form

$(x-h)^2+(y-k)^2=r^2$

With center $(h,k)$ and radius $r$ . In our example, we have

$(x+4)^2+(y-1)^2=9$

This tells us that we have a center at $(-4,1)$ , and a radius of $3$ .

To think about graphing the radius, a radius of $3$ is just the distance from the center of the circle to any endpoint.

After we interpret this information, we get the following graph:

graph{(x+4)^2+(y-1)^2=9 [-10, 10, -5, 5]}

Hope this helps!

How do you graph (x+4)^2+(y-1)^2=9(x+4)^2+(y-1)^2=9?