Finding the Center of a CircleDate: 06/06/99 at 15:37:38 From: Eddie Subject: Circle: given in form: Ax^2+Cy^2+Dx+Ey+F=0, find the center and radius and graph it. Hi, I am in advanced math in Rockville Centre, Long Island, and I need to know how to find the center and radius of a circle that is in the form: Ax^2 + Cy^2 + Dx + Ey + F = 0. It would be great if you could show me a sample problem or two. Thank you very much. Date: 06/07/99 at 11:59:17 From: Doctor Rick Subject: Re: Circle: given in form: Ax^2+Cy^2+Dx+Ey+F=0, find the center and radius and graph it. Hi, Eddie. The form you have given is actually more general than a circle; it is the form for an ellipse with axes parallel to the coordinate axes. You can see the equations on our Analytic Geometry Formulas FAQ: http://mathforum.org/dr.math/faq/formulas/faq.analygeom_2.html#twoellipses Look down that page for the circle formulas. A circle will have A = C. To find the center and radius, you want to put the equation into the standard form (x-h)^2 + (y-k)^2 = r^2 The center is then (h, k) and the radius is r. Here is an example: 5x^2 + 5y^2 + 10x + 6y + 6 = 0 Divide through by 5 first, so that the coefficient of x^2 and y^2 is 1: x^2 + y^2 + 2x + (6/5)y + 6/5 = 0 Group the terms in x and y separately: (x^2 + 2x) + (y^2 + (6/5)y) + 6/5 = 0 Complete the square in each set of parentheses. If you need help with this, you can search the Dr. Math Archives for completing the square. ((x + 1)^2 - 1) + ((y + 3/5)^2 - 9/25) + 6/5 = 0 Collect the constants and move them to the right: (x + 1)^2 + (y + 3/5)^2 = 4/25 Now you can read off the center: (-1, -3/5). The radius is the square root of the constant on the right, which is 2/5. Graph the circle; you will see that it passes through (-1, -1). Check this against the original equation: 5(-1)^2 + 5(-1)^) + 10(-1) + 6(-1) + 6 = 0 5 + 5 - 10 - 6 + 6 = 0 0 = 0 Good, this point is indeed on the circle. I hope this example has helped you. If you don't understand why I did any of these steps, write back and ask! - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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