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CIRCLES page 156 LINGKARAN halaman 156. Standar Kompetensi Menyusun persamaan lingkaran dan garis singgungnya.

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Presentation on theme: "CIRCLES page 156 LINGKARAN halaman 156. Standar Kompetensi Menyusun persamaan lingkaran dan garis singgungnya."— Presentation transcript:

1 CIRCLES page 156 LINGKARAN halaman 156

2 Standar Kompetensi Menyusun persamaan lingkaran dan garis singgungnya.

3 Kompetensi Dasar 3.1. Menyusun pers. lingkaran yg memenuhi persyaratan yg ditentukan - Merumuskan pers. lingkaran berpusat di (0, 0) dan (a, b). - Menentukan pusat dan jari-jari lingkaran yg pers.nya diketahui. - Menentukan pers. lingkaran yg memenuhi kriteria tertentu Menentukan pers. garis singgung pd lingkaran dalam berbagai situasi - Melukis garis yg menyinggung lingkaran dan menentukan sifatnya - Merumuskan pers. garis singgung yg melalui suatu titik pd lingk. - Merumuskan pers. garis singgung yg gradiennya diketahui.

4 3.1. Menyusun pers. lingkaran yg memenuhi persyaratan yg ditentukan - Merumuskan pers. lingkaran berpusat di (0, 0) dan (a, b). - Menentukan pusat dan jari-jari lingkaran yg pers.nya diketahui. - Menentukan pers. lingkaran yg memenuhi kriteria tertentu.

5 Jadwal Ulangan Lingkaran 1Lingkaran 2 11 IPA 1Kamis, 18 NovKamis, 2 Des 11 IPA 2Selasa, 16 NovSelasa, 30 Nov 11 IPA3Jumat, 19 NovRabu, 1 Des

6 Just what you need  Distance between 2 points  Gradient (m) A (x A, y A ) B(x B, y B ) xx yy

7  Linear (line) equation  Distance between point & line  → ax + by = ab M(p, q)  b a x y  r Change it into: ax + by + c = 0 Just what you need

8 Exp. 1 Calculate the distance between (2, 7) & (–1, 3) and its line gradient. Answer:  x = 2 – (–1) = 3  y = 7 – 3 = 4 Exp. 2 Find the line equation through point (6, 0) & (0, 8) Answer: 8x + 6y = 48 4x + 3y = Exp. 3 Find the distance (r) between point (2,–5) and line 4x – 3y = 12 Answer: Change 4x – 3y = 12 into 4x – 3y – 12 = 0 Point (2, –5)

9 CIRCLE EQUATION r x y r Circle equation with center (0, 0) is : x 2 + y 2 = r 2 Exp. 4 Find circle equation which its center is (0, 0) and has radius of 3 units. Answer: The equation is: x 2 + y 2 = 9

10 Position of a point to the circle Point J lies inside the circle so we said that: m 2 + n 2 < r 2 Point S is on the circle Point F is outside the circle x y r  F SS J  (m, n)

11 Do Exercisespage 161 no:1 b 2 b 4 b 6 a

12 CIRCLE EQUATION Circle equation with center P(a, b) is : (x–a) 2 + (y–b) 2 = r 2 Exp. 5 Find circle equation which its center is (2, 3) and has radius of 5 units. Answer: The equation is: (x–2) 2 + (y–3) 2 = 5 2 or x 2 + y 2 – 4x – 6y – 12 = 0  P a b r

13 Exp. 6 Find the circle equation which its center lies on P(2, 3) and the circle passes through point (–1, 4) Answer: First, calculate the radius The equation: (x – 2) 2 + (y – 3) 2 = 10 x 2 + y 2 – 4x – 6y + 3 = 0 A circle has diameter line that passes through (–5, 1) and (3, –7). Find its equation. Answer: Exp. 7  P P (–1, 4)  (–5, 1)   ( 3, –7)  P First, find the coordinate of P Second, find the radius (x – (–1)) 2 + (y – (– 3)) 2 = 32 x 2 + y 2 + 2x + 6y – 22 = 0

14 ANOTHER FORM OF CIRCLE EQUATION In the last equation: (x–a) 2 + (y–b) 2 = r 2 with center P(a, b) and radius r another form is: x 2 + y 2 + Ax + By + C = 0 with center and radius the coefficient of x 2 and y 2 must be 1

15 Exp. 8 Find the center of circle: x 2 + y 2 + 4x – 6y + 10 = 0 Answer: A = 4 B = –6 Center: P(–A/2, –B/2) P(–4/2, –(–6)/2) = (–2, 3) Exp. 9 A circle of x 2 + y 2 – 4x + 2y + C = 0 passes through (5, –1). Find its radius. Answer: First, find the value of C then insert (5, –1) into the equation (–1) 2 – (–1) + C = 0 C = –4

16 Exp. 10 Find the center and radius of: 2x 2 + 2y x – 12y + 6 = 0 Answer: First, divide it by 2 x 2 + y 2 + 8x – 6y + 3 = 0 Exp. 11 Circle x 2 + y 2 – 4x + 6y + k = 0 has a radius of 5 units. Find the value of k. Answer:

17 Do Exercises page 164 no:1 b2 b 34 b 5 b6 b 7 page 167 no:1 b2 b 3 b 58 9 Daily test 8: Circles 1

18 Next KD Menentukan pers. garis singgung pd lingkaran dlm berbagai situasi - Melukis garis yg menyinggung lingkaran & menentukan sifatnya - Merumuskan pers. garis singgung yg melalui suatu titik pd lingk. - Merumuskan pers. garis singgung yg gradiennya diketahui.

19 TANGENT LINE ON A CIRCLE Tangent (line) is a line that touch the circle in one point only. EE  F  F  G  H    P r r Line  touches the circle at point E, then line  is a tangent of the circle. Line  also a tangent of the circle. Line is not a tangent of the circle, because it intersects the circle at two points.

20 PROPERTIES OF A TANGENT A circle has so many tangents depend on the position of the point. If the point lies inside the circle (point K), then there is no tangent, because the line must be intersects the circle at 2 points. If the point lies on the circle exactly (point T), then there is 1 tangent only. If the point lies outside the circle (point D), then there are 2 tangents.  P K  K   T D  How do we know the position of the point to the circle?  just insert the point coordinate into the circle equation.

21 FINDING THE TANGENTS EQUATION Case 1: Point on the circle  P (a, b)  T (x 1, y 1 )  r STEPS: » Find the gradient of line PT (m PT ) » Because of line PT is perpendicular to line  then: m PT. m 1 = –1 » Use line equation formula to find the tangent equation: y – y 1 = m 1 (x – x 1 ) » Find the circle center

22 Exp. 12 Find the tangent line on circle (x – 2) 2 + (y + 1) 2 = 16 that passes through H(5, –1) Answer:  P (2, –1) From the graph above, let’s check the position of point H(5, –1) (x – 2) 2 + (y + 1) 2 = 16 H(5, –1) (5 – 2) 2 + (–1 + 1) 2 = 9 < 16 There is no tangent line. Exp. 13 Find the tangent line on circle x 2 + y 2 – 8x + 6y – 20 = 0 that passes through J(1, –9) Answer: Check the position of point J(1, –9) (–9) 2 – (–9) – 20 = ? – 8 – 54 – 20 = 0  The center: Gradient PJ: Gradient of tangent: m PJ. m 1 = –1m 1 = –1/2 Tangent equation: y – y 1 = m 1 (x – x 1 ) y – (–9) = –1/2 (x – 1) 2y + x + 17 = 0 RALAT

23 Exp. 13 Find the tangent line on circle x 2 + y 2 – 8x + 6y – 20 = 0 that passes through J(1, –9) Answer: Check the position of point J(1, –9) (–9) 2 – (–9) – 20 = ? – 8 – 54 – 20 = 0  The center: Gradient PJ: Gradient of tangent: m PJ. m 1 = –1m 1 = –1/2 Tangent equation: y – y 1 = m 1 (x – x 1 ) y – (–9) = –1/2 (x – 1) 2y + x + 17 = 0 Exp. 14 Another way to find the tangent on circle x 2 + y 2 – 8x + 6y – 20 = 0 that passes through J(1, –9) Answer: Change x 2 + y 2 – 8x + 6y – 20 = 0 into form (x – a) 2 + (y – b) 2 = r 2 Insert J(1, –9) into the equation 2y + x + 17 = 0 x 2 – 8x y 2 + 6y + 9 = (x – 4) 2 + (y + 3) 2 = 45 Change it: (x 1 –4)(x–4) + (y 1 +3)(y+3) = 45 (1–4)(x–4) + (–9 +3)(y+3) = 45–3(x–4) –6(y+3) = 45 –3x + 12 – 6y – 18 = 45

24 Exp. 15 Find the tangent line on circle x 2 + y 2 = 65 that passes through G(8, 1) Answer: Check the position of point G(8, 1) = 65  (on the circle) x 2 + y 2 = 65  x. x + y. y = 65 insert G(8, 1) 8x + 1y = 65  8x + y = 65 Exp. 16 Find the tangent line on circle x 2 + y 2 – 8y = 37 that passes through Z(2, –3) Answer: Check the position of point Z(2, –3) (–3) 2 – 8(–3) = ? = 37  (on the circle) x 2 + y 2 – 8y = 37  x 1.x + (y–4) 2 = 53 insert Z(2, –3) 2 x + –7(y–4) = 53 2 – 7y = 25 RALAT

25 Do Exercises page 175 no: 1 b 2 b

26 FINDING THE TANGENTS EQUATION Case 2: the gradient is known  P (a, b)  T (x 1, y 1 )  r STEPS: » Find the radius (r) » The tangent equation is: » Find the circle center P(a, b)  : because there will be 2 tangent lines (one parallel to another)

27 Exp. 17 A circle has equation: (x – 5) 2 + y 2 = 49 Find its tangent equation if the gradient is 3. Answer:  P(5, 0) 1 3 Exp. 18 If line y = x + k touches circle x 2 + y 2 = 25 then find the value of k. Answer: x 2 + y 2 = 25  P(0, 0) and r = 5 y – 0 = 1 (x – 0) + k y = x + k  m = 1 k 2 = r 2 (1 + m 2 ) = 5 2 ( ) = 50

28 FINDING THE TANGENTS EQUATION Case 3: the point is outside the circle  P (a, b) T ( x 1, y 1 )  STEPS » Find y 1 and y 2 and the two coordinates (x 1, y 1 ) and (x 2, y 2 ) » Insert each coordinate into circle equation x 1 x + y 1 y = r 2 to find the tangents equation. » Insert the polar equation into circle equation to find x 1 and x 2 » Find the polar line equation by changing x 2 + y 2 = r 2 aaaainto x 1 x + y 1 y = r 2

29 Exp. 19 Find the tangents line to the circle x 2 + y 2 = 25 that passes through (7, 1) Answer: Do Exercises page 180 no: 1 c 2 c 5 Polar: x 1 x + y 1 y = 25 7x + y = 25  y = 25 – 7x x 2 + y 2 = 25  x 2 + (25 – 7x) 2 = 25 x = 3  y = 4  (3, 4) x = 4  y = –3  (4, –3) (3, 4)  3x + 4y = 25 (4, –3)  4x – 3y = 25 (7, 1) lies outside the circle x 2 – 7x + 12 = 0  (x – 3)(x – 4) = 0 Exp. 20 Find the tangents line to the circle x 2 + y 2 – 2x – 6y – 15 = 0 that passes through point (6, –2) Answer: (6, –2) lies outside the circle Polar: (x – 1) 2 + (y – 3) 2 = 35 (6 – 1)(x – 1) + (–2 – 3)(y – 3) = 35 5(x – 1) – 5(y – 3) = 35 5x – 5y + 10 = 35  y = x – 5 (x – 1) 2 + (x – 5 – 3) 2 = 35 x 2 – 9x + 15 = 0  use abc formula RALAT

30 Do Exercises page 183 and from buku Mandiri page 56 – 70 FINISH


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