1 / 13

Higher

Higher. Circle. Unit 2 Outcome 4. The General equation of a circle. x 2 + y 2 + 2gx + 2fy + c = 0. Wednesday, 07 January 2009. Higher. Circle. Unit 2 Outcome 4. x 2 + y 2 + 2gx + 2fy + c = 0. In the same way we can. The equation of a circle is

Télécharger la présentation

Higher

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Higher Circle Unit 2 Outcome 4 The General equation of a circle x 2 + y 2 + 2gx + 2fy + c = 0 Wednesday, 07 January 2009

  2. Higher Circle Unit 2 Outcome 4 x 2 + y 2 + 2gx + 2fy + c = 0 In the same way we can The equation of a circle is (x – 2)2 + (y – 3)2 = 25 Write the equation without brackets (x – a) 2 + (y – b) 2 = r2 (x – a)(x – a) + (y – b)(y – b) = r2 (x – 2)(x – 2) + (y – 3)(y – 3) = 25 x2 – 2ax + a2 + y2 - 2by + b2 = r2 x2 - 4x + 4 + y2 - 6y + 9 = 25 x2 + y2 – 2ax – 2by +a2 +b2 – r2 = 0 x2 - 4x + y2 - 6y + 13 - 25 = 0 As a , b and r are constants (numbers) then these can be collected together as one term, c x2 + y2 - 4x - 6y - 12 = 0 x2 + y2 – 2ax – 2by + c = 0 This is the general form of the equation of a circle Wednesday, 07 January 2009

  3. Higher Circle Unit 2 Outcome 4 Radiusr 2. Centre C(-g,-f) x 2 + y 2 + 2gx + 2fy + c = 0 Radiusr 1. Centre C(a,b) Wednesday, 07 January 2009

  4. Finding the centre and the radius Given the equation of a circle, we can find the coordinates of its centre and the length of its radius. For example: Find the centre and the radius of a circle with the equation (x– 2)2 + (y + 7)2 = 64 By comparing this to the general form of the equation of a circle of radius r centred on the point (a, b): (x– a)2 + (y– b)2 = r2 We can deduce that for the circle with equation (x– 2)2 + (y + 7)2 = 64 The centre is at the point (2, –7) and the radius is 8. Wednesday, 07 January 2009

  5. Finding the centre and the radius When the equationof a circle is given in the form x2+ y2– 2ax – 2by + c= 0 we can use the method of completing the square to write it in the form (x– a)2 + (y– b)2 = r2 For example: Find the centre and the radius of a circle with the equation x2+ y2+ 4x – 6y + 9= 0 Start by rearranging the equation so that the x terms and the y terms are together: x2+ 4x+y2– 6y + 9= 0 Wednesday, 07 January 2009

  6. Finding the centre and the radius x2+ 4x+y2– 6y + 9= 0 We can complete the square for the x terms and then for the y terms as follows: x2 + 4x = (x + 2)2– 4 y2– 6y = (y– 3)2– 9 The equation of the circle can now be written as: (x + 2)2– 4 + (y– 3)2– 9 + 9 = 0 (x + 2)2 + (y– 3)2= 4 (x + 2)2 + (y– 3)2= 22 The centre is at the point (–2, 3) and the radius is 2. Wednesday, 07 January 2009

  7. Higher Circle Unit 2 Outcome 4 x 2 + y 2 + 2gx + 2fy + c = 0 Alternative approach x2+ 4x+y2– 6y + 9= 0 Rearrange to get in the general form C is sum of all the constants x 2 + y 2 + 2gx + 2fy + c = 0 x2+y2 + 4x – 6y + 9= 0 2g= 4 2f= -6 c= 9 g= 2 f= -3 c= 9 r2 = 22 + - 32 - 9 (x + 2)2 + (y– 3)2= 22 r2 = g2 +f2 - c As before It therefore follows that Centre (-g, -f) The centre is at the point (–2, 3) and the radius is 2. Wednesday, 07 January 2009

  8. Higher Circle Unit 2 Outcome 4 x 2 + y 2 + 2gx + 2fy + c = 0 Show that the equation x2 + y2 - 6x + 2y - 71 = 0 represents a circle and find the centre and radius. x2 + y2 - 6x + 2y - 71 = 0 2g= -6 2f= 2 c= -71 r2 = g2 + f2 -c c= -71 g= -3 f= 1 r2 = 9 + 1 - -71 (x + 3)2 + (y– 1)2= 92 r2 = 81 This is now in the form (x-a)2 + (y-b)2 = r2 Centre (-g, -f) So represents a circle with centre (3,-1) and radius = 9 Wednesday, 06 January 2009

  9. Higher Circle Unit 2 Outcome 4 x 2 + y 2 + 2gx + 2fy + c = 0 Show that the equation x2 + y2 + 6x - 2y - 15 = 0 represents a circle and find the centre and radius. x2 + y2 + 6x - 2y - 15 = 0 2g= 6 2f= -2 c= -15 r2 = g2 + f2 -c c= -15 g= 3 f= -1 r2 = 9 + 1 - -15 (x - 3)2 + (y+ 1)2= 52 r2 = 25 Centre (-g, -f) This is now in the form (x-a)2 + (y-b)2 = r2 So represents a circle with centre (-3,1) and radius = 5 Wednesday, 06 January 2009

  10. Higher Circle Unit 2 Outcome 4 x 2 + y 2 + 2gx + 2fy + c = 0 Show that the equation x2 + y2 - 4x - 6y + 9 = 0 represents a circle and find the centre and radius. x2 + y2 - 4x - 6y + 9 = 0 2g= -4 2f= -6 c= 9 r2 = g2 + f2 -c c= 9 g= -2 f= -3 r2 = 4 + 9 - 9 (x + 2)2 + (y+ 3)2= 22 r2 = 4 Centre (-g, -f) This is now in the form (x-a)2 + (y-b)2 = r2 So represents a circle with centre (2,3) and radius = 2 Wednesday, 06 January 2009

  11. Higher Circle Unit 2 Outcome 4 x 2 + y 2 + 2gx + 2fy + c = 0 Show that the equation x2 + y2 + 2x + 8y + 1 = 0 represents a circle and find the centre and radius. x2 + y2 + 2x + 8y + 1 = 0 2g= 2 2f= 8 c= 1 r2 = g2 + f2 - c c= 1 g= 1 f= 4 r2 = 1 + 16 -1 (x - 1)2 + (y- 4)2= 42 r2 = 16 Centre (-g, -f) This is now in the form (x-a)2 + (y-b)2 = r2 So represents a circle with centre (-1,-4) and radius = 4 Wednesday, 06 January 2009

  12. Higher Circle Unit 2 Outcome 4 Centre C(a,b)and radiusr (x – a)2 + (y – b)2 = r2 Page 170 To build skills Complete Exercise 3A Q 1, Q2, Wednesday, 06 January 2009

  13. Higher Circle Unit 2 Outcome 4 What do you see ? Tuesday, 06 January 2009

More Related