Question

The circle below is centered at the point (3, 1) and has a radits of length 2.

Answer 1

Answer:

Step-by-step explanation:

$Step 1: The- equation- of -a- circle- with -center -at- (h,k)- and- a -radius- of- r- is:\\(x-h)^2+(y-k)^2=r^2\\In -this -question ,h=3, k=1 and r=2, so- after- substituting- into- above -formula- we- get:\\(x-3)^2+(y-1)^2=2^2\\(x-3)^2+(y-1)^2 = 4\\ x^2-6x+9+y^2 -2y+1-4=0\\x^2-6x+y^2-2y+6=0$