How do you graph $2(x-3)^2+2y^2=8$ ?

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Answer 1 · 2016-03-03T14:23:03

Reduce to the standard form $(x-3)^2+ y^2 = 2^2$ .
This represents the circle with center at (3. 0) and radius = 2.

The equation $(x-a)^2+(y-b)^2 = r^2$ represents the locus of the point that moves at a constant distance r from (a, b)..

How do you graph 2(x-3)^2+2y^2=82(x-3)^2+2y^2=8?