How do you find the equation of a circle center of (5,-2) and a radius of 2?

2 Answers
Apr 3, 2018

#(x-5)^2+(y+2)^2=4#

Explanation:

I'm sure there a better answer and explanation but from my little experience with geometry I know the following:
1. #x^2+y^2=r^2# is the equation of a circle center of (0,0) and a radius of r.
2. to move the center you just need to move the points of x and y.

so a movement of 5 points in the positive direction of the x axis is translated to
#(x-5)^2# instead of #(x^2)#
a movement of 2 in the negative direction of the y axis is translated to
#(y+2)^2# instead of #(y^2)#

which then lead to the equation
#(x-5)^2+(y+2)^2=2^2#

Apr 3, 2018

#(x-5)^2+(y+2)^2=4#

Explanation:

The generic equation of a circle with center #(h,k)# and radius #r# is #(x-h)^2+(y-k)^2=r^2#.

With a center of #(5,-2)# and a radius of #2#, you have
#(x-5)^2+(y-(-2)^2)=2^2#
or
#(x-5)^2+(y+2)^2=4#.