Uncovering the Geometry of x^2+y^2=25 & y=sqrt 25-x^2

  • Thread starter Thread starter jeahomgrajan
  • Start date Start date
  • Tags Tags
    Geometry
Click For Summary

Homework Help Overview

The discussion revolves around the geometric interpretation of the equations x² + y² = 25 and y = √(25 - x²), specifically how the first equation represents a circle and the second a semi-circle.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of solving x² + y² = 25 for y, noting that it yields both positive and negative square roots. There is confusion regarding the relationship between the equations and their graphical representations.

Discussion Status

Some participants provide insights into why x² + y² = r² describes a circle, while others question the interpretation of y = 5 - x as a semi-circle. The conversation includes attempts to clarify the distinction between the equations and their respective geometric shapes.

Contextual Notes

Participants are navigating through the definitions and properties of circles and semi-circles, with some confusion about the implications of square roots in the context of these equations.

jeahomgrajan
Messages
49
Reaction score
0

Homework Statement



This is a simple math equation and i am a bit confused on how

x^2+y^2=25 makes a circle and how y= sqrt 25-x^2 makes a semi circle
 
Physics news on Phys.org
If you would solve x^{2} + y^{2} = 25 for y, you would not only get y=\sqrt{25-x^{2}}, but also something else. What?
 
y=5-x
 
jeahomgrajan said:
y=5-x
No, not quite. (Keep in mind that \sqrt{a^{2}-b^{2}}\neq a-b.)

Think of something as simple as e.g. x^{2}=9. There are two different values of x that satisfy this. What values would that be?
 
+-3?
 
Well, since the circle is defined as the locus of all points in a plane which are all a fixed and equal distance from the center, it is fairly intuitively obvious that the implicit equation maps to a circle. A more general case of the formula you presented is x^2 + y^2 = r^2 where r is the radius of the circle. Perhaps a general way to derive and illustrate how that equation makes a circle is to examine the circular definition of the trigonometric functions. We know that sin(theta) = y/r and cos(theta) = x/r where r is the radius and (x,y) is the coordinate of the radius' intersection with the circle. I'm also presuming you're familiar with the identity sin^2(theta) + cos^2(theta) = 1. Placing our unit circle definitions in place of the identity yields x^2 + y^2 = r^2 which may intuitively illustrate how it makes a circle.

Probably a better way of thinking about it is this: given any ordered pair (x,y) satisfying x^2 + y^2 = r^2, (x,y) must always be some fixed distance r from the center; hence, the equation produces the locus of all points on a plane an equal distance from the center. In other words, x^2 + y^2 = r^2 maps to a circle.
 
intence, but okay, what about the second equation which i have mentioned
 
jeahomgrajan said:
y=5-x ( that would make a semi circle right)

y = 5-x is a line, how are you getting semicircle
 
y = 5 - x will definitely not make a semi-circle. Solve x^2 + y^2 = r^2 for y and it should be fairly clear why it produces a semi-circle if x^2 + y^2 = r^2 creates a circle.
 
  • #10
jeahomgrajan said:
+-3?

Exactly. So if you now have y^{2}=25-x^{2}, what two values of y do you get?
 
  • #11
y= SqaUREROOT 25-x^2, (this should be a semicircle right?
 
  • #12
Yes, this is the upper half of the circle. The lower half is y = -\sqrt{25 - x^2}
 
  • #13
What people were trying to tell you was that if a = b^2 then b = \pm \sqrt{a} NOT just \sqrt{a}
 
  • #14
Alright i understand thanks
 

Similar threads

Replies
19
Views
2K
  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 15 ·
Replies
15
Views
4K
Replies
10
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
2K
Replies
13
Views
2K
Replies
3
Views
2K