How do you write an equation for a circle with center (2,0) and radius sqrt11?

1 Answer
Apr 24, 2016

Answer:

#x^2-4x+y^2-7=0#

Explanation:

The equation of a circle is

#x^2+y^2=r^2#,

where #r# is the radius, given as #sqrt11#, so

#x^2+y^2=sqrt11^2=11#.

You want to get the circle centered around the origin #(0,0)#, which you do by adding or subtracting a certain amount to the #x# and #y# values of the center to get it to #0#.

#x# is given as #2# (the point is #(2,0)#), so you subtract #2#. #y# is #0# so you can leave it the same. This gives an equation of

#(x-2)^2+y^2=11#.

Expanding this out,

#x^2-4x+4+y^2=11#

Rearranging for the final answer,

#x^2-4x+y^2-7=0#