What is the equation of the circle with a center at #(3 ,1 )# and a radius of #1 #?
1 Answer
Jan 20, 2016
Explanation:
The general form for the equation of a circle with a center at
#(x-h)^2+(y-r)^2=r^2#
We know that
#(h,k)rarr(3,1)=>h=3,k=1#
#r=1#
So the equation of the circle is
#(x-3)^2+(y-1)^2=1^2#
or, slightly more simplified (squaring the
#(x-3)^2+(y-1)^2=1#
The circle graphed:
graph{((x-3)^2+(y-1)^2-1)((x-3)^2+(y-1)^2-.003)=0 [-2.007, 9.093, -1.096, 4.454]}
