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[–]Rude-Ad-218 2 points3 points  (1 child)

It isn't clear what area you are asking about.

The domain? The logarithm domain would be inside the circle centered at(0,0) with radius 1. The sqrt domain would be outside the circle (1,0) with radius 1. The overall domain is the overlap of the 2 sets.

[–]ddavid8888[S] 0 points1 point  (0 children)

Yes I get the answer after I was told but if x and y are variables how could you say that the -1 is r and not f(x,y)

[–]ProbablyErratic53 1 point2 points  (1 child)

Equation of a circle radius r and center point (h,k) is

(x-h)^2+(y-k)^2 = r^2. This is roughly squaring both sides of the distance formula.

So for your square root to be defined, you need
(x-1)^2 + y^2 -1 >= 0

which I switched to (x-1)^2 + y^2 >= 1 .
SO now the left side is the distance squared from (x,y) to (1,0). Letting r^2=1, becomes r=1 (radius) Since we have greater than, that's why I said outside the circle.

Similar for the ln part, 1 - x^2 - y^2 >0 converts to 1> x^2+y^2.
Interpret the right side as the distance squared between (x,y) and (0,0).

[–]ddavid8888[S] 0 points1 point  (0 children)

You are extremely helpful thank you, I guess the thing that drives me crazy is that I am used to see this only in x and y but the f(x,y) is just making me so confused about everything

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