Asked by Knights
Find the largest real number x for which there exists a real number y such that x^2+y^2 = 2x+2y .
I think it is a circle, but how am i supposed to figure this out??
I think it is a circle, but how am i supposed to figure this out??
Answers
Answered by
bobpursley
The standard equation for a circle is ...
(x-a)^2 + (y-b)^2=r^2
Lets play with your equation.
x^2+y^2-2x-2y=0
x^2-2x + 1 + y^2-2y+1= 2 That wont work. We do have perfect squares on the left, but there is needed a zero on the right. Hmmm It is not a circle.
Lets look at if y is y<=0
if y is zero, then
x^2=2x or x can be any real number up to infinity, and there can be no larger x. So the answer is x=inf is the largest real number. Yes, y=0 is a real number.
(x-a)^2 + (y-b)^2=r^2
Lets play with your equation.
x^2+y^2-2x-2y=0
x^2-2x + 1 + y^2-2y+1= 2 That wont work. We do have perfect squares on the left, but there is needed a zero on the right. Hmmm It is not a circle.
Lets look at if y is y<=0
if y is zero, then
x^2=2x or x can be any real number up to infinity, and there can be no larger x. So the answer is x=inf is the largest real number. Yes, y=0 is a real number.
Answered by
Slater
@bobpursley your wrong.
x^2=2x does not mean x can be any number. If x was 4, then it would be 4^2=2*4, which simplifies too 16=8. THATS NOT TRUE! I think the anwser to this question is when x is 3, and y is one, the largest x = 3
x^2=2x does not mean x can be any number. If x was 4, then it would be 4^2=2*4, which simplifies too 16=8. THATS NOT TRUE! I think the anwser to this question is when x is 3, and y is one, the largest x = 3
Answered by
Slater
actually the anwser is sqrt(2) +1, my prevouis anwser is incorrect
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