Conics: Circles: Working with Equations (page 2 of 3) Sections: Introduction & Drawing, Working with equations, Further examples
The numerical side, the 16, is the square of the radius, so it actually indicates 16 = r^{2} = 4^{2}, so the radius is r = 4. Reading from the squaredvariable parts, the center is at (h, k) = (2, 3).
The numerical side tells me that r^{2} = 25, so r = 5. The xsquared part is really (x – 0)^{2}, so h = 0. The temptation is to read off the "3" from the ysquared part and conclude that k is 3, but this is wrong. The centervertex form has subtraction in it, so I need to convert first to that form. y + 3 = y – (–3) So the ycoordinate of the center is actually k = –3. radius r = 5, center (h, k) = (0, –3) Warning: It is very easy to forget that sign in the middle of the squared parts. Don't be careless! In the previous examples, information was extracted from a given equation. You'll also need to be able to work from given information backwards to find an equation.
I'll just plug the center and radius into the centerradius form: (x – (4))^{2}
+ (y – (–2))^{2} = 10^{2}
Since no particular form of the equation was specified, the above is an acceptable answer. If your book specifies some other format, then you may need to multiply things out: x^{2} –
8x + 16 + y^{2} + 4y + 4 = 100
Keep in mind that there is no standard meaning to the term "standard form". If your book specifies some other form, memorize that form for your tests.
To convert to centerradius form, I'll need to complete the squares. x^{2} + y^{2}
+ 2x + 8y + 8 = 0
Then the center is at (h, k) = (–1, –4) and the radius is r = 3.
The Distance Formula will give me the radius, which is, after all, the distance between the center and any point on the circle. Then the centerradius form of the equation is: (x – (–5))^{2}
+ (y – (12))^{2} = 5^{2}
The radius will be half the length of the diameter, and the midpoint of the diameter will be the center. The Midpoint Formula gives me: I can use either end of the diameter for my point on the circle; the distance between the center and the circle will be the same, regardless. I like smaller numbers, so I'll pick (–1, 0). The Distance Formula gives me: This distance is the length of the radius r, so r^{2} = 29. Plugging my results into the centerradius form, I get: Copyright © Elizabeth Stapel 20102011 All Rights Reserved (x – (4))^{2}
+ (y – (–2))^{2} = 29
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