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Frequently Asked Questions (FAQ)
What is x^2+2x*3+3^2+y^2-2y*2/3+(2/3)^2=4^2 ?
- The solution to x^2+2x*3+3^2+y^2-2y*2/3+(2/3)^2=4^2 is Circle with (a,b)=(-3, 2/3),r=4