# How the?

1. Feb 3, 2009

### jeahomgrajan

1. The problem statement, all variables and given/known data

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

2. Feb 3, 2009

### AssyriaQ

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?

3. Feb 3, 2009

y=5-x

4. Feb 3, 2009

### AssyriaQ

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?

5. Feb 3, 2009

+-3?

6. Feb 3, 2009

### jgens

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.

7. Feb 3, 2009

### jeahomgrajan

intence, but okay, what about the second equation which i have mentioned

8. Feb 3, 2009

### NoMoreExams

y = 5-x is a line, how are you getting semicircle

9. Feb 3, 2009

### jgens

y = 5 - x will definately 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. Feb 3, 2009

### AssyriaQ

Exactly. So if you now have $$y^{2}=25-x^{2}$$, what two values of y do you get?

11. Feb 4, 2009

### jeahomgrajan

y= SqaUREROOT 25-x^2, (this should be a semicircle right?

12. Feb 4, 2009

### Staff: Mentor

Yes, this is the upper half of the circle. The lower half is y = $-\sqrt{25 - x^2}$

13. Feb 4, 2009

### NoMoreExams

What people were trying to tell you was that if $$a = b^2$$ then $$b = \pm \sqrt{a}$$ NOT just $$\sqrt{a}$$

14. Feb 4, 2009

### jeahomgrajan

Alright i understand thanks