Question

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

Answer 1

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`

Answer 2

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 )