Set of points specified by x^2 + y^2 <= 4x + 4y

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The inequality x^2 + y^2 ≤ 4x + 4y describes a set of points within and on the boundary of a circle centered at (2, 2) with a radius of 2√2. By completing the square, the expression can be rewritten as (x-2)^2 + (y-2)^2 ≤ 8, indicating that all points satisfying this inequality lie inside or on the circle. The discussion clarifies that the original form suggests a circle centered at the origin, but the transformation reveals the correct center. The distinction between the inequality and the equation is emphasized, with the inequality representing all points inside the circle and the equation representing points on the circle itself. Understanding this transformation is crucial for correctly interpreting the geometric implications of the inequality.
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Homework Statement



What set of points is specified by the inequality x^2 + y^2 ≤ 4x + 4y

Homework Equations



x^2 + y^2 = r^2 is the formula for a circle with its center at the origin

The Attempt at a Solution



x^2 - 4x + y^2 - 4y ≤ 0 ?The book gives the solution, if you want me to post it i can. But i didn't understand how they got it
 
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nickadams said:

Homework Statement



What set of points is specified by the equation x^2 + y^2 ≤ 4x + 4y
This is actually an inequality, not an equation.
nickadams said:

Homework Equations



x^2 + y^2 = r^2 is the formula for a circle with its center at the origin

The Attempt at a Solution



x^2 - 4x + y^2 - 4y ≤ 0 ?


The book gives the solution, if you want me to post it i can. But i didn't understand how they got it

Complete the square in the x terms and in the y terms. The < part of the inequality represents all of the points inside a circle. The = part represents all the points on the circle.
 
Mark44 said:
This is actually an inequality, not an equation.

ninja edited :wink:

Complete the square in the x terms and in the y terms. The < part of the inequality represents all of the points inside a circle. The = part represents all the points on the circle.

are we completing the square to get the inequality in the familiar (x-a)^2 + (y-a)^2 = R^2 form that represents a circle? So it would be (x-2)^2 + (y-2)^2 ≤ 8

So apparently that means the same thing as (x-0)^2 + (y-0)^2 ≤ 4x + 4y, even though (x-0)^2 + (y-0)^2 would indicate that the circle is at the origin whereas (x-2)^2 + (y-2)^2 is for a circle with the center at (2,2).

Could someone help me understand why setting (x-0)^2 + (y-0)^2 less than or equal to "4x + 4y" rather than a number can make the circle centered at (2,2) instead of the origin as (x-0) and (y-0) led me to believe?
 
nickadams said:
ninja edited :wink:



are we completing the square to get the inequality in the familiar (x-a)^2 + (y-a)^2 = R^2 form that represents a circle? So it would be (x-2)^2 + (y-2)^2 ≤ 8
Yes. This inequality can be separated into two statements:
(x-2)^2 + (y-2)^2 < 8
(x-2)^2 + (y-2)^2 = 8
The inequality represents all the point inside the circle.
The equation represent all the points on the circle.

Together, the ≤ represents all the points on the circle or inside it.
nickadams said:
So apparently that means the same thing as (x-0)^2 + (y-0)^2 ≤ 4x + 4y,
Yes, but this form is not helpful at all.
nickadams said:
even though (x-0)^2 + (y-0)^2 would indicate that the circle is at the origin whereas (x-2)^2 + (y-2)^2 is for a circle with the center at (2,2).

Could someone help me understand why setting (x-0)^2 + (y-0)^2 less than or equal to "4x + 4y" rather than a number can make the circle centered at (2,2) instead of the origin as (x-0) and (y-0) led me to believe?
 
Mark44 said:
Yes. This inequality can be separated into two statements:
(x-2)^2 + (y-2)^2 < 8
(x-2)^2 + (y-2)^2 = 8
The inequality represents all the point inside the circle.
The equation represent all the points on the circle.

Together, the ≤ represents all the points on the circle or inside it.
Yes, but this form is not helpful at all.


Thanks Mark44!
 

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