# Math question (equation of circle )

1. Dec 23, 2004

### roger

Hello

Please could anyone explain this to me :

for the equation of circle :

(x-h)^2 + (y-k)^2 = r^2

I don't understand why the centre of the circle is at the point (h,k) ?

e.g

if (x-2)^2 + (y-k)^2 = 36

(8,0) is a point on the circle ,but 8 is obviously longer than the radius which is 6 ?

Thanks for any help

Roger

2. Dec 23, 2004

### quasar987

I guess you forgot to specify k = 0. In which case, yes, (8,0) is a point of the circle. The definition of a circle is "all points an equal distance from its center". This distance, called the radius, is specified to be 6 in your problem. But like you said the point (8,0) is on the circle, which means it must be a distance 6 from its center... aren't you forced to conclude that x = 2, y = 0 is the coordinate of it's center?

3. Dec 23, 2004

### futb0l

It just moves the circle. It's like saying:

$$y = (x-2)^2$$

is moving y = x^2 graph to the right by 2.

4. Dec 23, 2004

### HallsofIvy

Staff Emeritus
The distance from point (a, b) to point (x, y) is \radic;((x-a)2+ (y- b)2)

Since a circle is defined by the property that "the distance from any point on the circle to the center is a constant: the radius", if (x,y) is a point on a circle with center (a,b) and radius r, it must be true that \radic;((x-a)2+ (y- b)2)= r or, squaring both sides to get rid of the squareroot, (x-a)2+ (y- b)2= r2.

"if (x-2)^2 + (y-0)^2 = 36

(8,0) is a point on the circle ,but 8 is obviously longer than the radius which is 6 ?"
(I replace "k" by "0")
Yes, thus proving that (0,0) is NOT the center of the circle. The distance from the point (8,0) to (2, 0) is 8-2= 6 which is the radius.

5. Dec 23, 2004

### quasar987

More generally now... As you may have remarked, the simplest equation of the circle, x^2 + y^2 = r^2 ressembles very much pythagoras' square triangle equation. In fact, the equation of the circle is nothing but that. We say: "A circle of radius r is all sets of points (x,y) that satisfy pythagoras' equation for a given distance r".

In the simplest equation of the circle, x^2 represents the squared distance of the x coordinate from the origin and y^2 represents the squared distance of the y coordinate from the origin. So all you have to do to extent this idea to a circle not centered at the origin is remark that (x - h)^2 represents the squared distance of the x coordinate to the point (h,0) and that (y - k)^2 represents the squared distance of the x coordinate to the point (0,k).

Another way of seing it (perhaps easier?) is vectorially. Consider the point-vectors $\vec{a}=(h,k)$ and $\vec{b}=(x,y)$. You can easily convince yourself that $\vec{d} = \vec{b}-\vec{a}=(x-h,y-k)$ is a vector going from the point (h,k) to the point (x,y). Then if we set the restriction that the norm of the vector (which is also the distance from (h,k) to (x,y)) be equal to r, then we have, by the definition of the norm of a vector,

$$\sqrt{(x-h)^2+(y-k)^2}=r \Leftrightarrow (x-h)^2+(y-k)^2=r^2$$

which is yet another way to illustrate mathematically the intuitive definition of the circle: "All points an equal distance from its center."

(Does anybody know what this "bug" in LaTeX is due to or how to fix it? I'm talking about the wide useless space after vector b.)

Last edited: Dec 23, 2004
6. Dec 23, 2004

### roger

Thankyou for the help everyone.....

just one more query :

when the square root is taken, in this example, why don't we add the plus minus sign ?

Would it also depend on the definition of the square root ?

7. Dec 23, 2004

### quasar987

I don't know what square root you are talking about exactly but for exemple, in

$$r^2 = (x-h)^2 + (y-k)^2$$,

if we take the square root of both sides to get the expression of r, we discard the negative square root by convention. The radius, or more generally, a distance, is positive by definition.