General Form of equation for a Circle

In summary, the conversation discusses finding the radius of a circle given its equation in general form. The correct procedure involves completing the square and using the formula r=sqrt((a/2)^2+(b/2)^2-c). Another way to derive this formula is by comparing it to the equation of a circle with radius r and center point (x_0, y_0). Both methods yield the same result.
  • #1
Saladsamurai
3,020
7
I was helping my girlfriend earlier with some HW and now I don't have her text (since she is not here) but I have a question regarding circles.
I did not cover a lot of this material in precalc. and only recall the Standard Form of an equation: x^2+y^2=r^2.

So my question is: if given the equation in General Form, and I wish to know the radius, is the following procedure (completing the square) correct?

Given x^2+y^2+ax+by+c=0

x^2+ax+y^2+by+c=o
x^2+ax+(a/2)^2+y^2+by+(b/2)^2+c=(a/2)^2+(b/2)^2
(x+a/2)^2+(y+b/2)^2=(a/2)^2+(b/2)^2-c

so r=sqrt((a/2)^2+(b/2)^2-c)

is this correct?
thanks,
Casey
 
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  • #2
Yep, it is correct - and a nice way to do it too.

Another way to derive this is as follows: the equation of a circle with radius r and center point ([itex]x_0[/itex], [itex]y_0[/itex]) is [tex](x - x_0)^2 + (y - y_0)^2 = r^2[/tex]. Working out the brackets gives [tex]x^2 - 2 x_0 x + x_0^2 + y^2 - 2 y_0 y + y_0^2 - r^2 = 0[/tex]. Now if we compare to your formula [tex]x^2 + y^2 + a x + b y + c = 0[/tex] we read off that
[tex]a = -2 x_0, b = 2 y_0, c = x_0^2 + y_0^2 - r^2[/tex]
whence solving for r gives
[tex] r^2 = x_0^2 + y_0^2 - c [/tex] and you get the same formula (you can substitute -a/2 for [itex]x_0[/itex] and drop the minus since it's squared anyway, and similarly for -b/2).
 
  • #3
Nice. Thank you CompuChip.
 

1. What is the general form of an equation for a circle?

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

2. How do you determine the center and radius of a circle from its equation?

To determine the center and radius of a circle from its equation, you can rewrite the equation in the general form (x - h)^2 + (y - k)^2 = r^2. The values of h and k will represent the x and y coordinates of the center, and the value of r will represent the radius.

3. Can the general form of a circle's equation be written in other forms?

Yes, the general form of a circle's equation can also be written as (x - a)^2 + (y - b)^2 = c, where (a,b) represents the center of the circle and c represents the squared radius.

4. How is the general form of a circle's equation derived?

The general form of a circle's equation is derived using the Pythagorean theorem and the distance formula. By setting the distance between any point (x,y) on the circle and the center (h,k) equal to the radius r, we can create the equation (x - h)^2 + (y - k)^2 = r^2.

5. How is the general form of a circle's equation used in real-world applications?

The general form of a circle's equation is used in various fields such as geometry, physics, and engineering to model and solve problems involving circles. It is also used in computer graphics to create and manipulate circular objects.

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