Center and Radius of a circle
- Thread starter frichar3
- Start date
- Joined
- Feb 4, 2004
- Messages
- 16,582
"The program"? I'm sorry, but I don't know what this means.
Meanwhile, you need to complete the squares. That means putting the stuff on the left-hand side into (x - h)<sup>2</sup> and (y - k)<sup>2</sup> form, so you can read off the center (h, k) from the equation. Also, you need to simplify the values on the right-hand side. You have two numbers; add them.
Eliz.
Meanwhile, you need to complete the squares. That means putting the stuff on the left-hand side into (x - h)<sup>2</sup> and (y - k)<sup>2</sup> form, so you can read off the center (h, k) from the equation. Also, you need to simplify the values on the right-hand side. You have two numbers; add them.
Eliz.
- Joined
- Apr 12, 2005
- Messages
- 11,339
Let's try that again, from here.tkhunny said:
You have correctly, (1/2)<sup>2</sup> and 1<sup>2</sup>. Your notation does not say that. -1/2<sup>2</sup> is not the same as (-1/2)<sup>2</sup>. You need to decide what you mean and make sure that is what you write.
You also overlooked the constants that were factored out. See those 5s? You can't skip them. The values you add INSIDE the parentheses are multiplied by those constants. This gives:
5(x^2 - x + (1/2)<sup>2</sup>) + 5(y^2 + 2y + 1<sup>2</sup>) +3 - 5*(1/2)<sup>2</sup> - 5*1<sup>2</sup> = 0
It seems I faked you out a little by putting addition in front of the blanks at the end. You must add or subtract as necessary so that the value of the expression is not changed.
Now, let's see what you can do with it.