# CIRCLES page 156 LINGKARAN halaman 156

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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 IPA3 Jumat, 19 Nov Rabu, 1 Des

Just what you need  Distance between 2 points  Gradient (m)
(xA, yA) B(xB, yB) x y

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

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

CIRCLE EQUATION Circle equation with center (0, 0) is : x2 + y2 = r2
Exp. 4 Find circle equation which its center is (0, 0) and has radius of 3 units. Answer: The equation is: x2 + y2 = 9 r

Position of a point to the circle
x y r Point J lies inside the circle so we said that: m2 + n2 < r2  F S J  (m, n) Point S is on the circle Point F is outside the circle

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

CIRCLE EQUATION Circle equation with center P(a, b) is :
(x–a)2 + (y–b)2 = r2  P a b r 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 = 52 or x2 + y2 – 4x – 6y – 12 = 0

A circle has diameter line that passes through (–5, 1) and (3, –7).
Exp. 6 Exp. 7 A circle has diameter line that passes through (–5, 1) and (3, –7). Find its equation. Answer: Find the circle equation which its center lies on P(2, 3) and the circle passes through point (–1, 4) Answer:  P (–1, 4)  (–5, 1)   (3, –7) P First, calculate the radius First, find the coordinate of P Second, find the radius The equation: (x – 2)2 + (y – 3)2 = 10 (x – (–1))2 + (y – (– 3))2 = 32 x2 + y2 – 4x – 6y + 3 = 0 x2 + y2 + 2x + 6y – 22 = 0

ANOTHER FORM OF CIRCLE EQUATION
In the last equation: (x–a)2 + (y–b)2 = r2 with center P(a, b) and radius r another form is: x2 + y2 + Ax + By + C = 0 with center and radius the coefficient of x2 and y2 must be 1

Exp. 8 Exp. 9 Find the center of circle: x2 + y2 + 4x – 6y = 0 Answer: A circle of x2 + y2 – 4x + 2y + C = 0 passes through (5, –1). Find its radius. Answer: First, find the value of C A = 4 B = –6 then insert (5, –1) into the equation Center: P(–A/2, –B/2) 52 + (–1)2 – (–1) + C = 0 C = –4 P(–4/2, –(–6)/2) = (–2, 3)

Find the center and radius of: 2x2 + 2y2 + 16x – 12y + 6 = 0 Answer:
Exp. 10 Exp. 11 Find the center and radius of: 2x2 + 2y2 + 16x – 12y + 6 = 0 Answer: Circle x2 + y2 – 4x + 6y + k = has a radius of 5 units. Find the value of k. Answer: First, divide it by 2 x2 + y2 + 8x – 6y + 3 = 0

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

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. E  F G H P 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.

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. P K T D 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. 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: P (a, b)  T (x1, y1) r » Find the circle center » Find the gradient of line PT (mPT) » Because of line PT is perpendicular to line  then: mPT . m1 = –1 » Use line equation formula to find the tangent equation: y – y1 = m1 (x – x1)

Tangent equation: y – y1 = m1 (x – x1)
Exp. 12 Exp. 13 Find the tangent line on circle (x – 2)2 + (y + 1)2 = that passes through H(5, –1) Answer: Find the tangent line on circle x2 + y2 – 8x + 6y – 20 = that passes through J(1, –9) Answer: P (2, –1) Check the position of point J(1, –9) 12 + (–9)2 – (–9) – 20 = ? – 8 – 54 – 20 =  The center: Gradient PJ: From the graph above, let’s check the position of point H(5, –1) Gradient of tangent: RALAT (x – 2)2 + (y + 1)2 = 16 H(5, –1) mPJ . m1 = –1 m1 = –1/2 Tangent equation: y – y1 = m1 (x – x1) (5 – 2)2 + (–1 + 1)2 = 9 < 16 y – (–9) = –1/2 (x – 1) There is no tangent line. 2y + x = 0

Change x2 + y2 – 8x + 6y – 20 = 0 into form (x – a)2 + (y – b)2 = r2
Exp. 13 Exp. 14 Find the tangent line on circle x2 + y2 – 8x + 6y – 20 = that passes through J(1, –9) Answer: Another way to find the tangent on circle x2 + y2 – 8x + 6y – 20 = that passes through J(1, –9) Answer: Check the position of point J(1, –9) Change x2 + y2 – 8x + 6y – 20 = into form (x – a)2 + (y – b)2 = r2 12 + (–9)2 – (–9) – 20 = ? – 8 – 54 – 20 =  x2 – 8x y2 + 6y + 9 = The center: (x – 4)2 + (y + 3)2 = 45 Change it: (x1–4)(x–4) + (y1+3)(y+3) = 45 Gradient PJ: Insert J(1, –9) into the equation Gradient of tangent: mPJ . m1 = –1 m1 = –1/2 (1–4)(x–4) + (–9 +3)(y+3) = 45 Tangent equation: y – y1 = m1 (x – x1) –3(x–4) –6(y+3) = 45 y – (–9) = –1/2 (x – 1) –3x + 12 – 6y – 18 = 45 2y + x = 0 2y + x = 0

Exp. 15 Exp. 16 Find the tangent line on circle x2 + y2 = that passes through G(8, 1) Answer: Find the tangent line on circle x2 + y2 – 8y = that passes through Z(2, –3) Answer: Check the position of point G(8, 1) Check the position of point Z(2, –3) = 65  (on the circle) 22 + (–3)2 – 8(–3) = ? = 37  (on the circle) x2 + y2 =  x . x + y . y = 65 x2 + y2 – 8y = 37  x1.x + (y–4)2 = 53 insert G(8, 1) RALAT insert Z(2, –3) 8x + 1y = 65  8x + y = 65 2 x + –7(y–4) = 53 2 – 7y = 25

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

FINDING THE TANGENTS EQUATION
Case 2: the gradient is known P (a, b)  T (x1, y1) r STEPS: » Find the circle center P(a, b) » Find the radius (r) » 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 + y2 = 49
Find its tangent equation if the gradient is 3. Answer: If line y = x + k touches circle x2 + y2 = 25 then find the value of k. Answer: x2 + y2 = 25  P(0, 0) and r = 5 1 3 y = x + k  m = 1 P(5, 0) y – 0 = 1 (x – 0) + k k2 = r2 (1 + m2) = 52 (1 + 12) = 50

FINDING THE TANGENTS EQUATION
Case 3: the point is outside the circle STEPS T(x1, y1) » Find the polar line equation by changing x2 + y2 = r aaaainto x1 x + y1 y = r2 P (a, b) » Insert the polar equation into circle equation to find x1 and x2 » Find y1 and y2 and the two coordinates (x1, y1) and (x2, y2) » Insert each coordinate into circle equation x1 x + y1 y = r to find the tangents equation.

(7, 1) lies outside the circle (6, –2) lies outside the circle
Exp. 19 Exp. 20 Find the tangents line to the circle x2 + y2 = 25 that passes through (7, 1) Answer: Find the tangents line to the circle x2 + y2 – 2x – 6y – 15 = that passes through point (6, –2) Answer: (7, 1) lies outside the circle (6, –2) lies outside the circle Polar: x1 x + y1 y = 25 7x + y = 25  y = 25 – 7x Polar: (x – 1)2 + (y – 3)2 = 35 (6 – 1)(x – 1) + (–2 – 3)(y – 3) = 35 x2 + y2 = 25  x2 + (25 – 7x)2 = 25 5(x – 1) – 5(y – 3) = 35 5x – 5y = 35  y = x – 5 RALAT x2 – 7x = 0  (x – 3)(x – 4) = 0 x = 3  y = 4  (3, 4) x = 4  y = –3  (4, –3) (x – 1)2 + (x – 5 – 3)2 = 35 x2 – 9x + 15 = 0  use abc formula (3, 4)  3x + 4y = 25 (4, –3)  4x – 3y = 25 Do Exercises page 180 no: 1 c 2 c 5

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