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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

Respuesta :

D is correct . i just took the test

we are given

center=(-1,6)

radius=3

now, we can use standard equation of circle

[tex] (x-h)^2+(y-k)^2=r^2 [/tex]

where

center=(h,k)

radius =r

so, we get

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

[tex] h=-1,k=6 [/tex]

radius =r=3

[tex] r=3 [/tex]

now, we can plug these values

and we get

[tex] (x+1)^2+(y-6)^2=3^2 [/tex]

[tex] (x+1)^2+(y-6)^2=9 [/tex]

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

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