Question

Write an equation of the circle whose radius is 3 and whose center is (-1,6).

a.(x - 1)2 + (y + 6)2 = 3
b.(x - 1)2 + (y + 6)2 = 9
c.(x + 1)2 + (y - 6)2 = 3
d.(x + 1)2 + (y - 6)2 = 9 submit

Answer 1

D is correct . i just took the test

Answer 2

we are given

center=(-1,6)

radius=3

now, we can use standard equation of circle

`(x-h)^2+(y-k)^2=r^2`

where

center=(h,k)

radius =r

so, we get

center=(h,k)=(-1,6)

`h=-1,k=6`

radius =r=3

`r=3`

now, we can plug these values

and we get

`(x+1)^2+(y-6)^2=3^2`

`(x+1)^2+(y-6)^2=9`

so, option-D............Answer