The document discusses circles and their equations. It defines a circle as all points in a plane that are a fixed distance from a fixed center point. This fixed distance is called the radius. The standard form of a circle equation is (x-h)2 + (y-k)2 = r2, where (h,k) are the coordinates of the center and r is the radius. It also discusses converting between the standard and general forms of a circle equation.

1. CIRCLE Bea Nelene A. Que

3. (x-h) 2 + (y-k) 2 = r2 center radius form of a circle or standard form The center radius form of the equation of a circle simply because it extents the coordinates ( h,k ) of the center and the radius r of the circle. This form is also called the standard form for the equation of a circle. NOTE: If the center is at the origin, then h=k=0. Hence the standard form reduces to the form: x 2 + y 2 =r 2 .

4. Equation of Circles Circle whose center is at the origin Circle whose center is at (h,k)Equation: (This will be referred to as the "center- radius form". It may also be referred to as "standard form".)Example: Circle with center (0,0), radius 4 Equation: Graph: Example: Circle with center (2,-5), radius 3 Graph:

5. EXAMPLE Find the center and radius of the circle with equation x2 - 4x + y2 - 6y + 9 = 0 SOLUTION In order to find the center and the radius of the circle, we first rewrite the given equation into the standard form as given above in the definition. Put all terms with x and x2 together and all terms with y and y2 together using brackets. (x2 - 4x) +( y2 - 6y) + 9 = 0 We now complete the square within each bracket.. (x2 - 4x + 4) - 4 + ( y2 - 6y + 9) - 9 + 9 =0 (x - 2)2 + ( y - 3)2 - 4 - 9 + 9 = 0 Simplify and write in standard form (x - 2)2 + ( y - 3)2 = 4 (x - 2)2 + ( y - 3)2 = 22 We now compare this equation and the standard equation to obtain. center at C(h , k) = C(2 , 3) and radius r = 2

6. THE GENERAL FORM Consider again the standard form for the equation of a circle, i.e. (x-h) 2 + (y-k) 2= r2 Expanding the binomials and rearranging terms, we obtain x2 + y2 +(-2h)x +(-2h)y+(h2 + k2 -r2 ) =0 If we let D=-2h, E=-2k, and F=h2 +k2 =r2 then the preceding equation takes form.

8. When the equation of a circle appears in "general form", it is often beneficial to convert the equation to "center-radius" form to easily read the center coordinates and the radius for graphing.1. Convert into center-radius form. We will be creating two perfect square trinomials within the equation. • Start by grouping the x related terms together and the y related terms together. Move any numerical constants (plain numbers) to the other side.• Get ready to insert the needed values for creating the perfect square trinomials. Remember to balance both sides of the equation.• Find each missing value by taking half of the "middle term" and squaring. This value will always be positive as a result of the squaring process.• Rewrite in factored form. You can now read that the center of the circle is at (2, 3) and the radius is

9. To determine the graph of convert this equation to form (x-h) 2 + (y-k) 2 = r2 . If r2 > 0,then the graph is a circle with center. If r2= 0,then the graph is the single point. If r2< 0,then the graph is the null set. EXAMPLE: Determine whether the graph of the equation x 2 + y2 - 10 x - 8y + 32 = 0 is a circle, a point, or the null set.