How do you find the center and radius of # (x+3)^2 +y^2 = 37#?

2 Answers
Feb 3, 2016

Answer:

I found: center at #(-3,0)# and radius #6.1#.

Explanation:

We can compare our equation with the general equation of a circle with center at #(h,k)# and radius #r#:
#(x-h)^2+(y-k)^2=r^2#
giving us:
#h=-3#
#k=0#
#r=sqrt(37)~~6.1#

Feb 3, 2016

Answer:

centre (-3 , 0 ) , radius = #sqrt37 #

Explanation:

the standard form of the equation of a circle is

# (x-a)^2 + (y-b)^2 = r^2 #

where centre = (a , b ) and r = radius.

the equation here is in this form and so values of a , b and r
can be written down.

here a = -3 , b = 0 and # r^2 = 37 rArr r = sqrt37 #

hence centre = (-3 , 0 )