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a circle has the equation 2(x-2)^2+2y^2=2.?
a circle has the equation 2(x-2)^2+2y^2=2. find the center (h,k) and radius r and graph. Find the intercepts.
I really need help on this please.
1 Answer
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- William BLv 79 years agoFavorite Answer
you can divide through by 2 and get:
(x-2)^2 +y^2 =1
the center is (2,0) and the radius=1
when y=0,
(x-2)^2 =1
x-2=1, x-2=-1
x=3, x=1
the x-intercepts are (3,0) and (1,0)
when x=0,
(-2)^2 +y^2 =1
y^2=-3
there are no y-intercepts
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