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??

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.

@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

actually the anwser is sqrt(2) +1, my prevouis anwser is incorrect

To find the largest real number x for which there exists a real number y such that x^2 + y^2 = 2x + 2y, you can approach it using the concept of completing the square.

1. Start by moving all the terms to one side of the equation:
x^2 - 2x + y^2 - 2y = 0

2. Group the x-terms and the y-terms separately:
(x^2 - 2x) + (y^2 - 2y) = 0

3. Complete the square for the x-terms by adding the square of half the coefficient of x, which is (2/2)^2 = 1:
(x^2 - 2x + 1) + (y^2 - 2y) = 1

4. Similarly, complete the square for the y-terms by adding the square of half the coefficient of y, which is (2/2)^2 = 1:
(x^2 - 2x + 1) + (y^2 - 2y + 1) = 1 + 1

5. Simplify the equation:
(x - 1)^2 + (y - 1)^2 = 2

Now, the equation (x - 1)^2 + (y - 1)^2 = 2 represents a circle centered at (1, 1) with a radius of √2.

To find the largest real number x, you need to determine the rightmost point on the circle. Since the x-coordinate of the center is 1 and the radius is √2, adding the radius to the x-coordinate will give you the largest x-value. Therefore, the largest real number x is 1 + √2.

So, the largest real number x for which there exists a real number y such that x^2 + y^2 = 2x + 2y is 1 + √2.