How do you graph #x^2 + y^2 = 16#?

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

8

Answer:

This is going to be the graph of a circle with the center at the origin and a radius of 4.
graph{x^2+y^2-16=0 [-11.25, 11.25, -5.625, 5.625]}

Explanation:

The formula for a circle is:

#(x-x_c)^2 + (y-y_c)^2 = r^2#

The center of this circle will be located at the point (x_c, y_c).

You can expand your equation above to look like this:

#(x-0)^2 + (y-0)^2 = 16#

From this equation, you know that the center of the circle will be located at #(0;0)#.

Your radius can be found by using the equation as well. In your equation,
#r^2 = 16#
#:.sqrt(r^2) = sqrt(16)=> r = 4#.

The radius is 4.

Was this helpful? Let the contributor know!
1500