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!