Complete This Square: x^2 + y^2 - 6x + 2 = 0Date: 03/16/2003 at 23:40:35 From: Peter Subject: Completing THIS square: x^2 + y^2 - 6x + 2 = 0 Complete the square for: x^2 + y^2 - 6x + 2 = 0 It has a y variable, which makes it confusing; plus, I thought completing the square only involved quadratic functions, such as f(x) = ax^2 + bx + c = 0. This question relates to the equation of a circle; that is why I need to complete the square (to figure out its radius and loci/centre). What's the difference between a loci and centre again? Maybe you can answer that too please. :) Thanks. Oh, and if it helps, the centre is (3,0) and the radius is sqrt(7). Date: 03/17/2003 at 01:24:42 From: Doctor Jeremiah Subject: Re: Completing THIS square: x^2 + y^2 - 6x + 2 = 0 Hi Peter, When you have multiple variables you treat them as completely unrelated. In fact you can just ignore everything except the quadratic you are completing the square for. In this equation, the stuff in parentheses is the only important part: (x^2 - 6x + 2) + y^2 = 0 So all we really need to be concerned with is: (x^2 - 6x + 2) (x^2 - 6x + 2 + 7) - 7 (x^2 - 6x + 9) - 7 (x-3)^2 - 7 Now we have completed the square for the stuff in parentheses. We can substitute that back in: (x^2 - 6x + 2) + y^2 = 0 (x-3)^2 - 7 + y^2 = 0 (x-3)^2 + y^2 = 7 So this equation would have its center when x=3 and y=0 and the square of the radius is 7 so the radius is sqrt(7). Lets try another one. Start with: x^2 + y^2 - 16x + 12y + 19 = 0 Isolate the part you will complete the square for: (x^2 - 16x + 19) + y^2 + 12y = 0 (x^2 - 16x + 19 + 45) - 45 + y^2 + 12y = 0 (x^2 - 16x + 64) - 45 + y^2 + 12y = 0 (x-8)^2 - 45 + y^2 + 12y = 0 Now do the same for the other square: (x-8)^2 + (y^2 + 12y - 45) = 0 (x-8)^2 + (y^2 + 12y - 45 + 81) - 81 = 0 (x-8)^2 + (y^2 + 12y + 36) - 81 = 0 (x-8)^2 + (y+6)^2 - 81 = 0 Which gives us: (x-8)^2 + (y+6)^2 = 81 (x-8)^2 + (y+6)^2 = 9^2 As to your second question. http://www.mathworld.wolfram.com/ says a locus is "The set of all points (usually forming a curve or surface) satisfying some condition. For example, the locus of points in the plane equidistant from a given point is a circle, and the set of points in three-space equidistant from a given point is a sphere." And loci is the plural of locus. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/