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

[nickadams]

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

 

[Mark44]

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.

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.

 

[nickadams]

This is actually an inequality, not an equation.

ninja edited

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?

 

[Mark44]

ninja edited



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.

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.

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]

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!