CIRCLES page 156 LINGKARAN halaman 156 Standar Kompetensi

CIRCLES page 156 LINGKARAN halaman 156

Standar Kompetensi Menyusun persamaan lingkaran dan garis singgungnya.

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. 3. 2. 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.

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.

Jadwal Ulangan Lingkaran 1 Lingkaran 2 11 IPA 1 Kamis, 18 Nov Kamis, 2 Des 11 IPA 2 Selasa, 16 Nov Selasa, 30 Nov 11 IPA 3 Jumat, 19 Nov Rabu, 1 Des

Just what you need Distance between 2 points Gradient (m) B(x. B, y. B) y A (x. A, y. A) x

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

Exp. 1 Exp. 2 Exp. 3 Calculate the distance between (2, 7) & (– 1, 3) and its line gradient. Find the line equation through point (6, 0) & (0, 8) Find the distance (r) between point (2, – 5) and line 4 x – 3 y = 12 Answer: x = 2 – (– 1) = 3 Change 4 x – 3 y = 12 into 4 x – 3 y – 12 = 0 8 y = 7 – 3 = 4 Point (2, – 5) 6 8 x + 6 y = 48 4 x + 3 y = 24

CIRCLE EQUATION Circle equation with center (0, 0) is : x 2 + y 2 = r 2 r y x 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 r

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

Do Exercises no: 1 b 2 b 4 b 6 a page 161

CIRCLE EQUATION Circle equation with center P(a, b) is : (x–a)2 + (y–b)2 = r 2 r b P Exp. 5 Find circle equation which its center is (2, 3) and has radius of 5 units. Answer: a The equation is: (x– 2)2 + (y– 3)2 = 52 or x 2 + y 2 – 4 x – 6 y – 12 = 0

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

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

Exp. 8 Exp. 9 Find the center of circle: x 2 + y 2 + 4 x – 6 y + 10 = 0 A circle of x 2 + y 2 – 4 x + 2 y + C = 0 passes through (5, – 1). Find its radius. Answer: A=4 B = – 6 First, find the value of C then insert (5, – 1) into the equation Center: P(–A/2, –B/2) 52 + (– 1)2 – 4. 5 + 2 (– 1) + C = 0 C = – 4 P(– 4/2, –(– 6)/2) = (– 2, 3)

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

Do Exercises page 164 no: 1 b 2 b 3 5 b 6 b 7 1 b 2 b 3 b 5 8 9 page 167 no: Daily test 8: Circles 1 4 b

Next KD. . . 3. 2. 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.

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

PROPERTIES OF A TANGENT A circle has so many tangents depend on the position of the point. D K 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. P T If the point lies outside the circle (point D), then there are 2 tangents. How do we know the position of the point to the circle? just insert the point coordinate into the circle equation.

FINDING THE TANGENTS EQUATION Case 1: Point on the circle STEPS: » Find the circle center r P (a, b) T (x 1, y 1) » 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)

Exp. 12 Exp. 13 Find the tangent line on circle (x – 2)2 + (y + 1)2 = 16 that passes through H(5, – 1) Find the tangent line on circle x 2 + y 2 – 8 x + 6 y – 20 = 0 that passes through J(1, – 9) Answer: Check the position of point J(1, – 9) 12 + (– 9)2 – 8. 1 + 6 (– 9) – 20 = ? 1 + 81 – 8 – 54 – 20 = 0 P (2, – 1) The center: 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. RALAT Gradient PJ: Gradient of tangent: m. PJ. m 1 = – 1/2 Tangent equation: y – y 1 = m 1 (x – x 1) y – (– 9) = – 1/2 (x – 1) 2 y + x + 17 = 0

Exp. 13 Exp. 14 Find the tangent line on circle x 2 + y 2 – 8 x + 6 y – 20 = 0 that passes through J(1, – 9) Another way to find the tangent on circle x 2 + y 2 – 8 x + 6 y – 20 = 0 that passes through J(1, – 9) Answer: Check the position of point J(1, – 9) Change x 2 + y 2 – 8 x + 6 y – 20 = 0 12 + (– 9)2 – 8. 1 + 6 (– 9) – 20 = ? into form (x – a)2 + (y – b)2 = r 2 1 + 81 – 8 – 54 – 20 = 0 x 2 – 8 x + 16 + y 2 + 6 y + 9 = 20 + 16 + 9 The center: (x – 4)2 + (y + 3)2 = 45 Gradient PJ: Change it: (x 1– 4)(x– 4) + (y 1+3)(y+3) = 45 Gradient of tangent: Insert J(1, – 9) into the equation m. PJ. m 1 = – 1/2 Tangent equation: y – y 1 = m 1 (x – x 1) y – (– 9) = – 1/2 (x – 1) 2 y + x + 17 = 0 (1– 4)(x– 4) + (– 9 +3)(y+3) = 45 – 3(x– 4) – 6(y+3) = 45 – 3 x + 12 – 6 y – 18 = 45 2 y + x + 17 = 0

Exp. 15 Exp. 16 Find the tangent line on circle x 2 + y 2 = 65 that passes through G(8, 1) Find the tangent line on circle x 2 + y 2 – 8 y = 37 that passes through Z(2, – 3) Answer: Check the position of point G(8, 1) Check the position of point Z(2, – 3) 82 + 12 = 65 (on the circle) 22 + (– 3)2 – 8(– 3) = ? x 2 + y 2 = 65 x. x + y. y = 65 4 + 9 + 24 = 37 (on the circle) x 2 + y 2 – 8 y = 37 x 1. x + (y– 4)2 = 53 insert G(8, 1) insert Z(2, – 3) 8 x + 1 y = 65 8 x + y = 65 2 x + – 7(y– 4) = 53 2 – 7 y = 25 RALAT

Do Exercises page 175 no: 1 b 5 2 b 3 7 9 4

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

Exp. 17 Exp. 18 A circle has equation: (x – 5)2 + y 2 = 49 Find its tangent equation if the gradient is 3. 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 3 P(5, 0) 1 y =x+k m=1 y – 0 = 1 (x – 0) + k k 2 = r 2 (1 + m 2) = 52 (1 + 12) = 50

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

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

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