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

2 Answers
Mar 13, 2018

Answer:

This is a circle of radius #4# centred at the origin.

Explanation:

Given:

#x^2+y^2=16#

Note that we can rewrite this equation as:

#(x-0)^2+(y-0)^2 = 4^2#

This is in the standard form:

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

of a circle with centre #(h, k) = (0, 0)# and radius #r = 4#

So this is a circle of radius #4# centred at the origin:

graph{x^2+y^2 = 16 [-10, 10, -5, 5]}

Jul 7, 2018

Answer:

See below:

Explanation:

The equation of a circle is given by

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

With center #(h,k)# and radius #r#.

We have no #h# or #k# term, so we know our circle is centered at the origin.

We know from #sqrt16# that our radius is #4#. Now, we have everything we need to graph!

graph{x^2+y^2=16 [-10, 10, -5, 5]}

Hope this helps!