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How do you find the center and radius of the circle #(3x-1)^2 + (3y+3)^2 = -1#?

Answer 1

Elias Ackerman

This equation has no solutions in Real coordinates #(x, y)# so does not define a circle.

Note that if #x# and #y# are Real numbers then so are #3x-1# and #3y+3#, hence:

#(3x-1)^2 >= 0#

#(3y+3)^2 >= 0#

So #(3x-1)^2+(3y+3)^2 >= 0# cannot be equal to #-1#.

Since the equation has no solution, it defines an empty set of points, not a circle.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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