Maths Level 4 Module 8 hlayiso.com

Mathematics NQF Level 4 *see terms and conditions Downloaded from hlayiso.com Mathematics NQF Level 4

Space, shape and measurement Topic 3 Downloaded from hlayiso.com Mathematics NQF Level 4

Use the Cartesian coordinate system to derive and apply equations Module 8 Downloaded from hlayiso.com Mathematics NQF Level 4

Learning activity 8.1 Downloaded from hlayiso.com Mathematics NQF Level 4

1. Calculate the equation of a circle with its centre: 1.1) At the origin and with radius 3. 2 = [Equation x +y r 2 2 of circle with centre (0;0)] 2 2 2 3 ∴x + y = 2 2 9 ∴x + y = Downloaded from hlayiso.com Learning activity 8.1

1. Calculate the equation of a circle with its centre: 1.2) At (2;−1) and with radius 4. 2 2 ( x − a ) + ( y − b) = r 2 [Equation of circle with centre (a; b)] 2 2 2 ( x − 2) + ( y − (−1)) =4 2 2 ∴ x − 4x + 4 + y + 2 y +1 =16 2 2 ∴ x + y − 4 x + 2 y − 11 = 0 Downloaded from hlayiso.com Learning activity 8.1

Learning activity 8.2 Downloaded from hlayiso.com Mathematics NQF Level 4

x2 + y 2 − 4 x − 6 y + 9 =0 3. If the equation of a circle is and the centre of the circle is at (2;b), calculate b. [VIDEO] Click above to play this video. Downloaded from hlayiso.com Learning activity 8.2

7. Calculate the equation of the tangent to a circle at the point (5;8,732) if the centre of the circle is at (4; 7) and the radius of the circle is 2 units. Also calculate the equation of the circle. [VIDEO] Click above to play this video. Downloaded from hlayiso.com Learning activity 8.2

Summative assessment Module 8 Downloaded from hlayiso.com Mathematics NQF Level 4

1. Calculate the equation of a circle with its centre at (−2;4) and its radius 3. 2 2 2 ( x − a ) + ( y − b) = r ∴ ( x + 2) 2 + ( y − 4) 2 = 32 ∴ x 2 + 4 x + 4 + y 2 − 8 y + 16 =9 2 2 ∴ x + y + 4 x − 8 y + 11 = 0 [Equation of circle with centre (a; b)] Downloaded from hlayiso.com Summative assessment – Module 8

2. Calculate the equation of the tangent to x2 + y2 = 16 through (1;3,87). Method 1: 2 x1 x + y1 y r = (1) x + (3,87) y 16 = ∴ y =−0, 258 x + 4,134 Figure 8.6: Summative assessment Question 2 [Equation of tangent to circle with centre (0;0)] Downloaded from hlayiso.com Summative assessment – Module 8

2. Calculate the equation of the tangent to x2 + y2 = 16 through (1;3,87). (Continued) Method 2: yB − yA mradius = xB − xA 3,87 − 0 = 1− 0 = 3,87 Figure 8.6: Summative assessment Question 2 [Gradient of a straight line] Downloaded from hlayiso.com Summative assessment – Module 8

2. Calculate the equation of the tangent to x2 + y2 = 16 through (1;3,87). (Continued) Method 2: mradius ⋅ mTangent −1 = −1 ∴ mT = 3,87 = −0, 258 Figure 8.6: Summative assessment Question 2 [Perpendicular lines] Downloaded from hlayiso.com Summative assessment – Module 8

2. Calculate the equation of the tangent to x2 + y2 = 16 through (1;3,87). (Continued) Method 2: Equation of tangent: y − y1 = m( x − x1 ) y − 3,87 − = 0, 258( x − 1) y − = 0, 258 x + 0, 258 + 3,87 Figure 8.6: Summative assessment = −0, 258 x + 4,128 Question 2 Downloaded from hlayiso.com Summative assessment – Module 8

3. Calculate the equation of the tangent to the circle at the point (6;6) if the centre of the circle is at (2;3) and the radius is 5 units. Figure 8.7 Downloaded from hlayiso.com Summative assessment – Module 8

3. Calculate the equation of the tangent to the circle at the point (6;6) if the centre of the circle is at (2;3) and the radius is 5 units. [Equation of tangent to Method 1: circle with centre (a; b)] r2 ( x1 − a )( x − a ) + ( y1 − b)( y − b) = 52 (6 − 2)( x − 2) + (6 − 3)( y − 3) = 4x − 8 + 3y − 9 =25 4x + 3y = 42 3y =−4 x + 42 y − = 1,333 x + 14 Downloaded from hlayiso.com Summative assessment – Module 8

3. Calculate the equation of the tangent to the circle at the point (6;6) if the centre of the circle is at (2;3) and the radius is 5 units. Method 2: 6−3 y − y1 = m( x − x1 ) mr = 6−2 4 3 y − 6 =− ( x − 6) = 3 4 y =−1,333 x + 6 + 7,998 mr mT = −1 4 ∴ y =−1,333 x + 14 ∴ mT = − 3 Downloaded from hlayiso.com Summative assessment – Module 8

4. Find the equation of the circle that passes through the point (1;1), has a radius of 10, and of which the centre lies on the line y = 3x − 7. Let ( a; b ) be centre of circle ( x − a ) 2 + ( y − b) 2 = r2 ∴ ( x − 2) 2 + ( y − 3) 2 5 = 2 [Equation of circle] Figure 8.8 Downloaded from hlayiso.com Summative assessment – Module 8

4. Find the equation of the circle that passes through the point (1;1), has a radius of 10, and of which the centre lies on the line y = 3x − 7. = 3x − 7 y [Equation of radius] ∴ b = 3a − 7 ...(1) [Substitute point (a; b)] 2 2 [Equation of circle ( x − a ) + ( y − b) = 100 centre (a; b)] (1 − a ) 2 + (1 − b) 2 = 100 ...(2) [Substitute point (1;1)] Downloaded from hlayiso.com Summative assessment – Module 8

4. Find the equation of the circle that passes through the point (1;1), has a radius of 10, and of which the centre lies on the line y = 3x − 7. Substitute (1) into (2): (1 − a ) 2 + (1 − (3a − 7)) 2 = 100 1 − 2a + a 2 + [1 − 3a + 7]2 = 100 1 − 2a + a 2 + [8 − 3a ]2 = 100 2 2 1 − 2a + a + 64 − 48a + 9a = 100 2 10a − 50a − 35 = 0 Downloaded from hlayiso.com Summative assessment – Module 8

4. Find the equation of the circle that passes through the point (1;1), has a radius of 10, and of which the centre lies on the line y = 3x − 7. 2 ∴ 2a − 10a − 7 =0 10 ± 100 − 4(2)(−7) a= 2 10 ± 12,5 = 2 11, 25 ∴a = Downloaded from hlayiso.com Summative assessment – Module 8

4. Find the equation of the circle that passes through the point (1;1), has a radius of 10, and of which the centre lies on the line y = 3x − 7. Substitute into (1): = 3a − 7 ...(1) b = 3(11, 25) − 7 ∴b =26, 75 Downloaded from hlayiso.com Summative assessment – Module 8

4. Find the equation of the circle that passes through the point (1;1), has a radius of 10, and of which the centre lies on the line y = 3x − 7. Substitute a and b to find equation of circle: [Equation of circle centre (a; b)] ( x − a ) 2 + ( y − b) 2 = r2 ( x − 11, 25) 2 + ( y − 26, 75) 2 = 100 x 2 − 22,5 x + 12,56 y 2 + 126,56 − 53,5 y + 715,56 100 = x 2 + y 2 − 22,5 x − 53,5 y + 742,12 0 = Downloaded from hlayiso.com Summative assessment – Module 8

5. Find the equation of the tangent to the circle with equation x2 + y2 + 2x− 4y− 6 = 0 through the point (1 ; 4,646) x2 + y 2 + 2 x − 4 y − 6 =0 2 2 ( x + 2 x) + ( y − 4 y ) 6 = Complete the square 2 1 2 2 1 2 1 2 1 [ x + 2 x + ( .2) ] + [ y − 4 y + ( . − 4) ] =6 + ( .2) + ( . − 4) 2 2 2 2 2 ( x 2 + 2 x + 1) + ( y 2 − 4 y + 4) = 6 + 1 + 4 Downloaded from hlayiso.com Summative assessment – Module 8

5. Continued 2 ( x − a ) + ( y − b) = r2 2 [Equation of circle with centre (a; b)] 2 2 ( x + 1) + ( y − 2) = 11 Centre of circle is (−1; 2) r 2 = 11 ∴ r = 11 = 3,317 Radius is 3,317 Downloaded from hlayiso.com Summative assessment – Module 8

5. Find the equation of the tangent to the circle with equation x2 + y2 + 2x− 4y− 6 = 0 through the point (1 ; 4,646). Now we can find the equation of the tangent: Method 1: Using (1; 4, 646) and (−1; 2) y2 − y1 mr = x2 − x1 4, 646 − 2 = 1 − (−1) = 1,323 Downloaded Figure 8.9 from hlayiso.com Summative assessment – Module 8

5. Find the equation of the tangent to the circle with equation x2 + y2 + 2x− 4y− 6 = 0 through the point (1 ; 4,646). −1 mr ⋅ mT = Equation of tangent: −1 y − y1 = m( x − x1 ) mT = 1,323 y − 4, 646 − = 0, 756( x − 1) = −0, 756 ∴ y =−0, 756 x + 0, 756 + 4, 646 = −0, 756 x + 5, 402 Downloaded from hlayiso.com Summative assessment – Module 8

5. Find the equation of the tangent to the circle with equation x2 + y2 + 2x− 4y− 6 = 0 through the point (1 ; 4,646). Method 2: r2 ( x1 − a )( x − a ) + ( y1 − b)( y − b) = (1 − (−1))( x − (−1)) + (4, 646 − 2)( y − 2) =(3,317) 2 (2)( x + 1) + (2, 646)( y − 2) = 11, 002 2 x + 2 + 2, 646 y − 5, 292 = 11, 002 ∴ 2, 646 y =−2 x + 11,586 Downloadedyfrom = −0, 756 x + 5.402 hlayiso.com Summative assessment – Module 8

6. Calculate the equation of the tangent to the circle at the point (4;7852) if the centre of the circle is at (2; 3) and the radius is 5 units. ( x − a ) 2 + ( y − b) 2 = r2 ( x − 2) 2 + ( y − 3) 2 5 = 2 [Equation of circle] Figure 8.10 Downloaded from hlayiso.com Summative assessment – Module 8

6. Calculate the equation of the tangent to the circle at the point (4;7,582) if the centre of the circle is at (2; 3) and the radius is 5 units. Method 1: Equation of tangent at the point ( 4; 7,582 ) r2 ( x1 − a )( x − a ) + ( y1 − b)( y − b) = 52 ∴ (4 − 2)( x − 2) + (7,582 − 3)( y − 3) = ∴ 4 x − 8 − 2 x + 4 + 7,582 y − 22, 746 − 3 y + 9 = 25 ∴ 2 x − 4 + 4,582 y − 13, 746 =25 Downloaded from hlayiso.com Summative assessment – Module 8

6. Calculate the equation of the tangent to the circle at the point (4;7,582) if the centre of the circle is at (2; 3) and the radius is 5 units. ∴ 4,582 y = 42, 786 − 2 x y − = 0, 436 x + 9,33 Downloaded from hlayiso.com Summative assessment – Module 8

6. Calculate the equation of the tangent to the circle at the point (4;7,582) if the centre of the circle is at (2; 3) and the radius is 5 units. Method 2: y2 − y1 mr = x2 − x1 −1 7,582 − 3 mT = = 2, 291 4−2 = −0, 436 = 2, 291 Downloaded from hlayiso.com Summative assessment – Module 8

6. Calculate the equation of the tangent to the circle at the point (4;7,582) if the centre of the circle is at (2; 3) and the radius is 5 units. Equation of tangent at the point ( 4; 7,582 ) ( y − y1 ) = m( x − x1 ) y − 7,582 − = 0, 436( x − 4) y − = 0, 436 x + 1, 746 + 7,582 = −0, 436 x + 9,328 Downloaded from hlayiso.com Summative assessment – Module 8

7. Calculate the equation of the tangent to x2 + y2 + 6x− 8y− 25 = 0 at (2;9). 2 2 x + y + 6 x − 8 y − 25 = 0 ( x 2 + 6 x) + ( y 2 − 8 y ) = 25 Complete the square 1 2 1 1 2 1 [ x + 6 x + ( .6) ] + [ y − 8 y + ( . − 8) ] = 25 + ( .6) + ( . − 8) 2 2 2 2 2 2 2 2 2 2 ( x + 6 x + 9) + ( y − 8 y + 16) = 25 + 9 + 16 Downloaded from hlayiso.com Summative assessment – Module 8

7. Continued 2 ( x − a ) + ( y − b) = r2 2 [Equation of circle with centre (a; b)] 2 2 ( x + 3) + ( y − 4) = 50 2 Centre of circle is (−3; 4) r = 50 ∴ r = 50 = 7.07 Radius is 7, 07 Downloaded from hlayiso.com Summative assessment – Module 8

7. Calculate the equation of the tangent to x2 + y2 + 6x− 8y− 25 = 0 at (2;9). Now we can find the equation of the tangent: Method 1: Using (2;9) and (−3; 4) y2 − y1 9−4 mr = = x2 − x1 2 − (−3) 5 = 5 =1 Downloaded Figure 8.11: Question 7 from hlayiso.com Summative assessment – Module 8

7. Calculate the equation of the tangent to x2 + y2 + 6x− 8y− 25 = 0 at (2;9). −1 mr ⋅ mT = Equation of tangent: ∴ mT = (1) − 1 y − y1 = m( x − x1 ) = −1 y − 9 =−1( x − 2) y =− x + 2 + 9 ∴ y =− x + 11 Downloaded from hlayiso.com Summative assessment – Module 8

7. Calculate the equation of the tangent to x2 + y2 + 6x− 8y− 25 = 0 at (2;9). Method 2: r2 ( x1 − a )( x − a ) + ( y1 − b)( y − b) = (2 − (−3))( x + 3) + (9 − 4)( y − 4) =7, 072 5 x + 15 + 5 y − 20 = 49,985 49,985 + 20 − 15 ∴ y =− x + 5 y =− x + 11 Downloaded from hlayiso.com Summative assessment – Module 8

8. The centre of a circle lies on the line x− 2y− 1 = 0. The x-axis and the line y = 6 are tangents to the circle. Write down the equation of the circle. x-axis and y = 6 are tangents ∴ radius = 3 and centre is ( h;3) centre lies on x − 2 y − 1 =0 [Given] substitute (h;3) : x − 2 y − 1 =0 h − 2(3) − 1 =0 ∴h = 7 Downloaded from hlayiso.com Summative assessment – Module 8

8. The centre of a circle lies on the line x− 2y− 1 = 0. The x-axis and the line y = 6 are tangents to the circle. Write down the equation of the circle. Now we have centre ( 7;3) and r = 3 ( x − a ) 2 + ( y − b) 2 = r2 [Equation of circle with centre (a; b)] ∴ ( x − 7) 2 + ( y − 3) 2 9 = Downloaded from hlayiso.com Summative assessment – Module 8

9. Find the equation of the diameter of a circle with equation x2 + y2− 8x + 6y+ 21 = 0, which, when produced, passes through the point (2; 1). x 2 + y 2 − 8 x + 6 y + 21 = 0 ( x 2 − 8 x) + ( y 2 + 6 y ) = −21 ( x − 8x + 16 ) + ( y + 3 y + 9 ) =−21 + 16 + 9 2 2 2 2 ( x − 4) 2 + ( y + 3) 2 =−21 + 16 + 9 ( x − 4) 2 + ( y + 3) 2 = 4 [Complete the square] ∴ Centre is (4; −3) Downloaded from hlayiso.com Summative assessment – Module 8

9. Find the equation of the diameter of a circle with equation x2 + y2− 8x + 6y+ 21 = 0, which, when produced, passes through the point (2; 1). To find equation of diameter: Use point (2;1) and the centre (4; −3) y2 − y1 1 − −3 4 mdiameter = = = = −2 x2 − x1 2−4 −2 ∴ y =−2 x + c [Equation of straight line] Downloaded from hlayiso.com Summative assessment – Module 8

9. Find the equation of the diameter of a circle with equation x2 + y2− 8x + 6y+ 21 = 0, which, when produced, passes through the point (2; 1). Substitute point (2;1): 1 = 2(−2) + c ∴c =5 ∴ y =−2 x + 5 Downloaded from hlayiso.com Summative assessment – Module 8

Mathematics NQF Level 2 Downloaded from hlayiso.com Mathematics NQF Level 4