How do you graph #x^2 + y^2 = 81#?

2 Answers
Feb 22, 2016

Answer:

It is an equation of a circle with centre at #(0,0)# and radius #9#.

Explanation:

The equation is of the form of a circle with center at origin, as in the general form of a quadratic equation

#ax^2+2hxy+by^2+2fx+2gy+c=0#, while coefficients of #x^2# and #y^2# are equal (i.e. #a=b#), #f, g, h# are all zeros. In fact equation #x^2+y^2=81# graph{x^2+y^2=81 [-20, 20, -10, 10]} can be written as

#(x-0)^2+(y-0)^2=9^2#

and hence it is an equation of a circle with centre at #(0,0)# and radius #9#.

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#.

Since we have no #h# or #k# term, we know that we are centered at the origin.

From #sqrt81#, we know that our radius is #9#. Now we can graph!

graph{x^2+y^2=81 [-20, 20, -10, 10]}

Hope this helps!