You can put this solution on YOUR website! (I assume you want to simplify and/or interpret the equation. You should specify it when you state the problem and not make the tutors guess what you are looking for.)
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I like this problem because too often textbooks limit themselves to problems with artificially simple coefficients. Whether this problem comes from a real-world source or not, it shows the kinds of numbers you encounter in actual applications.
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Notice that this equation is quadratic in both x and y. In other words, there is both an term and a term. The process for this kind of problem is to group the x terms and y terms together and complete the square twice. .
We want to factor the x-terms into a single squared binomial, and do the same again for the y-terms. If we add the appropriate constant term in each case, each trinomial will be the square of a binominal and so can be factored.
The pattern for completing the square comes from .
Note that the constant term is half of the x-coefficient squared. This gives us the procedure: add half of the x (or y) coefficient squared, then factor. You also have to add the same number to the other side of the equation to keep it valid.
Doing this, we get:
Factoring the two trinomials on the left (one in x, one in y) yields: , which can be rewritten .
This is the equation of a circle centered on the point (4259.7,8225.7) with radius 237.
