How do you find the center and radius of # (x+3)^2 +y^2 = 37#?
I found: center at
We can compare our equation with the general equation of a circle with center at
centre (-3 , 0 ) , radius =
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 )