Math question (equation of circle )

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And also, the equation of a circle isr^2 = d(a,c),where r is the radius of the circle, and d(a,c) is the distance between the center of the circle and a point c on the circle. It is important to notice that r is never negative, it's a distance, and as such it is positive. But d(a,c) can be positive or negative, depending on the position of the point c relatively to the center of the circle. But that's another story...In summary, the equation of a circle is defined as "all points an equal distance from its center". This distance is represented by the radius, which is always positive. Therefore, when taking the square root to find
  • #1
roger
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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
 
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  • #2
roger said:
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 ?

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

[tex]y = (x-2)^2[/tex]

is moving y = x^2 graph to the right by 2.
 
  • #4
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
roger said:
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) ?

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 [itex]\vec{a}=(h,k)[/itex] and [itex]\vec{b}=(x,y)[/itex]. You can easily convince yourself that [itex]\vec{d} = \vec{b}-\vec{a}=(x-h,y-k)[/itex] 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,

[tex]\sqrt{(x-h)^2+(y-k)^2}=r \Leftrightarrow (x-h)^2+(y-k)^2=r^2[/tex]

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.)
 
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  • #6
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
I don't know what square root you are talking about exactly but for exemple, in

[tex]r^2 = (x-h)^2 + (y-k)^2[/tex],

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.
 

1. What is the equation of a circle?

The equation of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h,k) is the center of the circle and r is the radius.

2. How do I find the center and radius of a circle from its equation?

To find the center and radius, you can compare the equation of the circle to the standard form (x - h)^2 + (y - k)^2 = r^2. The value of h and k will give you the coordinates of the center, while the value of r will give you the radius.

3. Can I find the equation of a circle if I know three points on its circumference?

Yes, you can find the equation of a circle if you know three points on its circumference. You can use the formula (x - h)^2 + (y - k)^2 = r^2 and plug in the coordinates of the three points to form a system of equations. Solving this system will give you the values of h, k, and r, which you can then use to write the equation of the circle.

4. How do I graph a circle from its equation?

To graph a circle from its equation, first find the center and radius using the method mentioned in the second question. Then, plot the center point on the coordinate plane and use the radius to draw the circle. You can also plot a few more points on the circle to ensure accuracy.

5. Can I use the equation of a circle to find its area?

Yes, you can use the equation of a circle to find its area. The formula for the area of a circle is A = πr^2, where r is the radius. By substituting the value of r from the equation of the circle, you can find the area of the circle.

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