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# Circle

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### Circle

1. 1. CIRCLE<br />Bea Nelene A. Que<br />
2. 2. Center-radius form<br />A circle is the set S of all points in R2 that are at a fixed distance r from a fixed point C.<br />A circle is a locus(set) of points in a plain equidistant from a fixed point. <br /><ul><li>The fixed point C is called the center of the circle and the fixed distance r is called the radius. Note that r is the undirected distance from the center C to any point P on the circle. Hence, the value of r is a nonnegative real number or r>0. it thus becomes clear that a circle with center C and radius r is the graph of the relation.</li></li></ul><li>S= |P(x,y)|r=|CP| (1)<br />Let C (h,k) be the center of a circle and let P(x,y) be any point on the circle. Then from (1) and by the distance formula, we have:<br /> (x-h)2 + (y-k)2 = r2<br /> Consequently, we have proved the following theorem.<br /> A circle with center C(h,k) and the radius r is the graph of the relation.<br />
3. 3. (x-h) 2 + (y-k) 2 = r2 center radius form of a circle or standard form<br />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. <br />NOTE:<br />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 .<br />
4. 4. Equation of Circles<br />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: <br />Graph: Example:  Circle with center (2,-5), radius 3<br />   Graph:                 <br />
5. 5. EXAMPLE<br />Find the center and radius of the circle with equation<br />                                            x2 - 4x + y2 - 6y + 9 = 0<br />SOLUTION<br />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.<br />                                            (x2 - 4x) +( y2 - 6y) + 9 = 0<br />We now complete the square within each bracket.. <br />                                      (x2 - 4x + 4) - 4 + ( y2 - 6y + 9) - 9 + 9 =0                                          <br />                                       (x - 2)2  + ( y - 3)2 - 4 - 9 + 9 = 0<br />Simplify and write in standard form<br />                                              (x - 2)2  + ( y - 3)2 = 4  <br />                                               (x - 2)2  + ( y - 3)2 = 22<br />We now compare this equation and the standard equation to obtain.<br />                                            center at C(h , k) = C(2 , 3)<br />                                            and radius    r = 2<br />
6. 6. THE GENERAL FORM<br />Consider again the standard form for the equation of a circle, i.e.<br /> (x-h) 2 + (y-k) 2= r2 <br /> Expanding the binomials and rearranging terms, we obtain<br />x2 + y2 +(-2h)x +(-2h)y+(h2 + k2 -r2 ) =0<br /> If we let D=-2h, E=-2k, and F=h2 +k2 =r2 then the preceding equation takes form.<br />
7. 7. NOTE:<br /><ul><li> General form is an equation of the second degree in x and y (i.e. quadratic in x and y.)
8. 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.</li></ul>1. Convert   into center-radius form.<br />We will be creating two perfect square trinomials within the equation. <br />• 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.<br />You can now read that the center of the circle is at (2, 3) and the radius is <br />
9. 9. To determine the graph of convert this equation to form (x-h) 2 + (y-k) 2 = r2 .<br />If r2 > 0,then the graph is a circle with center.<br />If r2= 0,then the graph is the single point.<br />If r2< 0,then the graph is the null set.<br />EXAMPLE:<br />Determine whether the graph of the equation x 2 + y2 - 10 x - 8y + 32 = 0 is a circle, a point, or the null set.<br />
10. 10. SOLUTION:<br /> x 2 – y2 - 10x – 8y + 32 = 0 <br /> (x 2- 10x ) + (y2– 8y ) = -32<br /> (x 2 - 10x + 25) + (y 2 – 8y + 16) = -32 + 25 + 16<br /> (x- 5) 2 +(y- 4 ) 2 = 9<br /><ul><li>Since r2 = 9 > 0, then by the graph given, equation is a circle with center ( 5, 4 ) and the radius 3.</li></li></ul><li>THANKYOU…..<br />