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MAT 104 ELEMENTARY III 3 CREDITS

CO-ORDINATE (CONT’D)

CIRCLES The of a with (푎, 푏) and 푟 is given by (푥 − 푎)2 + (푦 − 푏)2 = 푟2 When the circle has its centre at origin, the equation of the circle becomes or reduces to

푥2 + 푦2 = 푟2

Because, circle with centre at origin means the coordinates of the centre is (0, 0) ⟹ (푥 − 푎)2 + (푦 − 푏)2 = 푟2

(푥 − 0)2 + (푦 − 0)2 = 푟2

푥2 + 푦2 = 푟2

Examples 1. Find the equation of the circle which has its center at origin and a radius of 5. Solution Simply from 푥2 + 푦2 = 푟2 ⟹ 푥2 + 푦2 = 52 Thus 푥2 + 푦2 = 25

2. Find the equation of the circle with center (2, 5) and a radius of 3. Solution From (푥 − 푎)2 + (푦 − 푏)2 = 푟2 ⟹ (푥 − 2)2 + (푦 − 5)2 = 32 ⟹ 푥2 + 푦2 − 4푥 − 10푦 + 20 = 0

3. Find the equation of a circle with centre (−2, 3) which passes through the (1, 2) Solution

Length of radius, 푟 = √(−2 − 1)2 + (3 − 2)2

푟 = √−32 + 12 푟 = √10 푟2 = 10

Using (푥 − 푎)2 + (푦 − 푏)2 with the circle centre (−2, 3) and radius 푟 = √10 The equation of the circle is (푥 + 2)2 + (푦 − 3)2 = 10 ⟹ 푥2 + 푦2 + 4푥 − 6푦 + 3 = 0

4. Find the radius and centre of the circle whose equation is given by 3푥2 + 3푦2 − 24푥 + 12푦 + 11 = 0 Solution 11 푥2 + 푦2 − 8푥 + 4푦 + = 0 3 11 푥2 + 푦2 − 8푥 + 4푦 = − 3 Completing 11 푥2 − 8푥 + 16 + 푦2 + 4푦 + 4 = − + 16 + 4 3

49 (푥 − 4)2 + (푦 + 2)2 = 3

49 Comparing with (푥 − 푎)2 + (푦 − 푏)2 = 푟2 ⟹ 푎 = 4, 푏 = −2, 푟 = √ 3 7 Circle centre (4, −2) and radius √3

EQUATION OF A CIRCLE GIVEN ABOUT THE DIAMETER The equation of a circle can also be deduced if the information about the points which the diameter passes through is known. In summary the equation of a circle whose diameter is the join of (푥1, 푥2) and (푦1, 푦2) is given by the equation

(푥 − 푥1)(푥 − 푥2) + (푦 − 푦1)(푦 − 푦2) = 0

Examples 1. Find the equation of a circle drawn on diameter 퐴퐵 where 퐴 is (−1, 3) and 퐵 is (3, 2) Solution 5 Mid-point of 퐴퐵; (1, ) ⟹ centre of circle 2 5 Thus, 푟2 = (1 + 1)2 + ( − 3)2 2 1 푟2 = 22 + (− )2 2 17 푟2 = 4 (Note that to get the length of 푟, you can use any of the two points with the 5 5 circle centre i.e. {(−1, 3) and (1, )} or {(3, 2) and (1, )} 2 2 5 17 Equation of circle (푥 − 1)2 + (푦 − )2 = 2 4 25 17 푥2 − 2푥 + 1 + 푦2 − 5푦 + = 4 4 푥2 + 푦2 − 2푥 − 5푦 + 3 = 0 Using the method

(푥 − 푥1)(푥 − 푥2) + (푦 − 푦1)(푦 − 푦2) = 0 ⟹ (푥 + 1)(푥 − 3) + (푦 − 3)(푦 − 2) = 0 ⟹ (푥2 − 2푥 − 3) + (푦2 − 5푦 + 6) = 0 ⟹ 푥2 + 푦2 − 2푥 − 5푦 + 3 = 0

2. The equation of a circle is 푥2 + 푦2 − 6푥 + 8푦 = 24. If the circle which passes through the point (−1, 1), find the coordinates of the other end of the diameter and the equation of the diameter.

Solution 푥2 + 푦2 − 6푥 + 8푦 = 24 푥2 − 6푥 + 9 + 푦2 + 8푦 + 16 = 24 + 9 + 16 (푥 − 3)2 + (푦 + 4)2 = 49

Thus the circle centre is (3, −4) ⟹ 푚푖푑푝표푖푛푡 표푓 푡ℎ푒 푑푖푎푚푒푡푒푟

Note that the first end of the diameter is (−1, 1), so the coordinates of the other end of the diameter is obtained as

1 1 3 = (푥 + 푥 ) −4 = (푦 + 푦 ) 2 1 2 2 1 2

1 1 3 = (−1 + 푥 ) −4 = (1 + 푦 ) 2 2 2 2

푥2 = 7 푦2 = −9

Thus the coordinates of the other end of the diameter is (7, −9).

Now since the coordinates of the two points of the diameter is known (−1, 1) and (7, −9). the equation of the (i.e. the diameter) is obtained using the two-point form

푦2− 푦1 푦 − 푦1 = (푥 − 푥1) 푥2− 푥1 −9 −1 푦 − 1 = (푥 + 1) 7+ 1

5 푦 − 1 = − (푥 + 1) ⟹ 4푦 + 5푥 + 9 = 0 4

Alternatively,

To find the equation of the diameter, we can use just the coordinates of the point that we were given i.e. (−1, 1) and the coordinates of the circle centre i.e. (3, −4), which is equally the mid-point of the diameter.

Using the two-point form,

푦2− 푦1 푦 − 푦1 = (푥 − 푥1) 푥2− 푥1 −4 −1 푦 − 1 = (푥 + 1) 3 + 1

5 푦 − 1 = − (푥 + 1) ⟹ 4푦 + 5푥 + 9 = 0 4

(The reason for this is because the three points – the two ends of the diameter and the circle centre i.e. (−1, 1), (7, −9) and (3, −4) lie on the same straight line and as such are collinear points meaning they will always have the same gradient).

GENERAL FORM OF THE EQUATION OF A CIRCLE The equation of a circle with centre (푎, 푏) and radius 푟 is (푥 − 푎)2 + (푦 − 푏)2 = 푟2 which gives 푥2 + 푦2 − 2푎푥 − 2푏푦 + 푎2 + 푏2 − 푟2 = 0 written in the form 푥2 + 푦2 + 2푔푥 + 2푓푦 + 푐 = 0, which is the general standard form of the equation of a circle. Any of 푔, 푓 and 푐 can be 0 With This general form of the equation of a circle, the equation of a circle can be determined given that it passes through three points by substituting the (푥, 푦) values of each point into the equation above and solving the three simultaneously.

Examples 1. Find the equation of a circle through the points (6, 1), (3,2) and (2, 3) Using 푥2 + 푦2 + 2푔푥 + 2푓푦 + 푐 = 0 For the point (6, 1), we have 62 + 12 + 2(6)푔 + 2(1)푓 + 푐 = 0 (푖) For the point (3, 2), we have 32 + 22 + 2(3)푔 + 2(2)푓 + 푐 = 0 (푖푖) For the point (2, 3), we have 22 + 32 + 2(2)푔 + 2(3)푓 + 푐 = 0 (푖푖푖) This becomes 37 + 12푔 + 2푓 + 푐 = 0 (푖푣) 13 + 6푔 + 4푓 + 푐 = 0 (푣) 13 + 4푔 + 6푓 + 푐 = 0 (푣푖)

Subtract equation (푣푖) from (푣), we have 2푔 − 2푓 = 0 ⟹ 푔 = 푓 (푣푖푖)

Subtract equation (푣) from (푖푣), we have 24 + 6푔 − 2푓 = 0 (푣푖푖푖)

Substitute for 푔 in equation (푣푖푖푖) 24 + 6푓 − 2푓 = 0 (since 푔 = 푓) 4푓 = −24 푓 = −6 ⟹ 푔 = −6 Solving for 푐, we have 푐 = 47 Thus the equation of the circle is 푥2 + 푦2 − 12푥 − 12푦 + 47 = 0

2. Find the equation of a circle through the points (0, 0), (3, 1) and (5, 5) Using 푥2 + 푦2 + 2푔푥 + 2푓푦 + 푐 = 0 For the point (0, 0), we have 02 + 02 + 2(0)푔 + 2(0)푓 + 푐 = 0 ⟹ 푐 = 0 (푖) For the point (3, 1), we have 32 + 12 + 2(3)푔 + 2(1)푓 + 푐 = 0 ⟹ 10 + 6푔 + 2푓 + 푐 = 0 (푖푖) For the point (5, 5), we have 52 + 52 + 2(5)푔 + 2(5)푓 + 푐 = 0 ⟹ 50 + 10푔 + 10푓 + 푐 = 0 (푖푖푖) Solve equations (푖푖) and (푖푖푖) simultaneously, we have 푔 = 0 and 푓 = −5 Thus the equation of the circle is 푥2 + 푦2 − 10푦 = 0

CONCYCLIC POINTS Four or more points are said to be concyclic if they lie on the same circle. Example Determine if the points (5, 2), (2, 3), (−3, −2) and (6, −5) are concyclic. Solution Using 푥2 + 푦2 + 2푔푥 + 2푓푦 + 푐 = 0 For the point 52 + 22 + 2(5)푔 + 2(2)푓 + 푐 = 0 (푖) For the point 22 + 32 + 2(2)푔 + 2(3)푓 + 푐 = 0 (푖푖) For the point −32 + −22 + 2(−3)푔 + 2(−2)푓 + 푐 = 0 (푖푖푖) Solving equations (푖), (푖푖) and (푖푖푖) simultaneously, 푔 = −2, 푓 = 2, 푐 = −17 Thus the equation of the circle is 푥2 + 푦2 − 4푥 + 4푦 − 17 = 0 Now if the points are concyclic input the coordinates of the fourth point into the equation of the circle to verify if this fourth point satisfies the equation Therefore, we put the point (6, −5) into the equation 푥2 + 푦2 − 4푥 + 4푦 − 17 = 0 We have, 62 + (−5)2 − 4(6) + 4(−5) − 17 = 0 ⟹ 36 + 25 − 24 − 20 − 17 = 0 This shows that the points are concyclic.

POINTS INSIDE OR OUTSIDE A CIRCLE Consider the 푓(푥, ) = (푥 − 푎)2 + (푦 − 푏)2 − 푟2. 푓(푥, 푦) = 0 defines a circle with radius 푟 and centre (푎, 푏). If the point (ℎ, 푘) lies inside the circle then (ℎ − 푎)2 + (푘 − 푏)2 < 푟2 i. e. (ℎ − 푎)2 + (푘 − 푏)2 − 푟2 < 0 or 푓(ℎ, 푘) < 0. So if (ℎ, 푘) lies inside the circle given by 푓(푥, 푦) = 0, then 푓(ℎ, 푘) < 0 and if the point (ℎ, 푘) lies outside the circle then 푓(ℎ, 푘) > 0. Example 1. Are the points 퐴 is (−1, −1) and 퐵 is (5, 2) inside or outside the circle 푥2 + 푦2 − 3푥 + 4푦 = 12 .

Let 푓(푥, 푦) = 푥2 + 푦2 − 3푥 + 4푦 − 12 푓(1, −1) = 1 + 1 − 3 − 4 − 12 = −17 < 0 ⟹ 퐴 is inside the circle

푓(5, 2) = 25 + 4 − 15 + 8 − 12 = 10 > 0 ⟹ 퐵 is outside the circle

POINT OF INTERSECTION OF A STRAIGHT LINE AND A CIRCLE

The coordinates of the point of intersection of a straight line and a circle is obtained by solving the equation of the line and the circle simultaneously.

EQUATION OF TANGENTS TO A CIRCLE (at the point (풙, 풚) on the circle) Tangents are drawn perpendicular to the radius of the circle from the point the radius touches the . To obtain the equation of the tangent, we differentiate the equation of the circle implicitly and thereafter substitute for the (푥, 푦) value in the result to arrive at the gradient, 푚. After getting the value of the gradient 푚, we then make use of the value of 푚 and the (푥, 푦) value to get the equation of the tangent to the circle using the 푔푟푎푑푖푒푛푡 − 표푛푒 푝표푖푛푡 푓표푟푚 method of finding the equation of a line.

Examples 1. Find the equation of the tangent to the circle 푥2 + 푦2 − 2푥 + 4푦 = 15 at the point (−1, 2)

Solution 푑푦 푑푦 Differentiating (implicitly), 2푥 + 2푦 − 2 + 4 = 0 푑푥 푑푥 푑푦 (2푥 + 4) = 2 − 2푥 푑푥 푑푦 2−2푥 = 푚 = 푑푥 2푦+4 푑푦 2−2(−1) 1 |(−1,2) = = = 푚 푑푥 2(2)+ 4 2 Therefore equation of the tangent is

푦 − 푦1 = 푚(푥 − 푥1) 1 푦 − 2 = (푥 + 1) 2 2푦 = 푥 + 5

2. Find the equation of the tangent to the circle 4푥2 + 4푦2 − 12푥 + 24푦 = 55 3 at the point (− , 1) 2 Solution 푑푦 푑푦 Differentiating (implicitly), 8푥 + 8푦 − 12 + 24 = 0 푑푥 푑푥 푑푦 (8푥 + 24) = 12 − 8푥 푑푥 푑푦 12−8푥 = 푑푥 8푦 + 24 3 푑푦 3 12−8(− ,) 24 3 | (− , 1) = 2 = = 푑푥 2 8(1)+ 24 32 4 Therefore equation of the tangent is

푦 − 푦1 = 푚(푥 − 푥1) 3 3 푦 − 1 = (푥 + ) 4 2 3 4푦 − 4 = 3(푥 + ) 2 8푦 − 8 = 6(푥 + 3) 8푦 − 6푥 = 26

AREA OF A

The of a triangle 퐴퐵퐶 whose vertices are 퐴(푥1, 푦1), 퐵(푥2, 푦2) and 퐶(푥3, 푦3) is given by 1 Area of ∆퐴퐵퐶 = |푥 푦 − 푥 푦 + 푥 푦 − 푥 푦 + 푥 푦 − 푥 푦 | 2 1 2 2 1 2 3 3 2 3 1 1 3

Examples 1. Find the area of the triangle 퐴퐵퐶, if (3, 4), 퐵(9, 7) and 퐶(5, −3) 1 Using the formula, ∆퐴퐵퐶 = |푥 푦 − 푥 푦 + 푥 푦 − 푥 푦 + 푥 푦 − 푥 푦 | 2 1 2 2 1 2 3 3 2 3 1 1 3 1 We have, |(3.7) − (9.4) + (9. −3) − (5.7) + (5.4) − (3. −3)| 2 1 1 th⟹ |21 − 36 − 27 − 35 + 20 + 9| = |−48| 2 2 1 ⟹ . 48 = 24 푠푞. 푢푛푖푡푠 2

ADDENDUM TO VECTORS Component of a vector in the direction of another vector

퐵⃗⃗ The component of a vector 퐴⃗ in the direction of vector 퐵⃗⃗ is given as 퐴⃗. |퐵⃗⃗| Examples

1. Let 퐴⃗ = 푖 + 4푗 and 퐵⃗⃗ = 3푖 + 4푗. Find the component of

i. 퐴⃗ in the direction of 퐵⃗⃗

ii. 퐵⃗⃗ in the direction of 퐴⃗ Solution

i. |퐵⃗⃗| = √32 + 42 = 5

퐵⃗⃗ 3푖+4푗 퐴⃗. = 푖 + 4푗 . |퐵⃗⃗| 5

(푖+4푗) .(3푖+4푗) 3+16 19 = ⟹ = 5 5 5

ii. |퐴⃗| = √12 + 42 = √17

퐴⃗ 푖+4푗 퐵⃗⃗. = 3푖 + 4푗 . |퐴⃗| √17

(3푖+4푗).(푖+4푗) 3+16 19 = ⟹ = 5 √17 √17 2. Find the value of 푎 such that the component of 퐴⃗ in the direction of vector 퐵⃗⃗

is equal to 0, if 퐴⃗ = 푎푖 + 2푗 and 퐵⃗⃗ = 푖 + 3푗. Solution

|퐵⃗⃗| = √12 + 32 = √10

퐵⃗⃗ 푖+3푗 퐴⃗. = 푎푖 + 2푗 . |퐵⃗⃗| √10 (푎푖+2푗) .(푖+3푗) 푎+6 = ⟹ √10 √10

퐵⃗⃗ 푎+6 Thus, 퐴⃗. = 0 ⟹ ⟹ 푎 + 6 = 0 |퐵⃗⃗| √10 Hence 푎 = −6

VECTOR PROJECTION

퐴⃗ .퐵⃗⃗ The scalar projection of a vector 퐴⃗ unto a vector 퐵⃗⃗ is given as Å⃗⃗⃗ = |퐵⃗⃗ |

While the vector projection of a vector 퐴⃗ unto a vector 퐵⃗⃗ is given as

퐴⃗ .퐵⃗⃗ 푝푟표푗 퐴⃗ = (퐵⃗⃗). 퐵⃗⃗ 퐵⃗⃗ .퐵⃗⃗

Examples ⃗⃗ ⃗⃗⃗ 1. Given the vector 퐴 = 5푖 − 푗 + 2푘 and 퐵 = 2푖 − 푗 − 3푘. Find (i) 푝푟표푗퐴⃗퐵 (ii) Å Solution

퐵⃗⃗ .퐴⃗ (i) 푝푟표푗 퐵⃗⃗ = (퐴⃗) 퐴⃗ 퐴⃗ .퐴⃗ 퐵⃗⃗ . 퐴⃗ = (2푖 − 푗 − 3푘). (5푖 − 푗 + 2푘) = 10 + 1 − 6 = 5 퐴⃗ . 퐴⃗ = (5푖 − 푗 + 2푘). (5푖 − 푗 + 2푘) = 25 + 1 + 4 = 30 5 5 1 1 푝푟표푗 퐵⃗⃗ = (5푖 − 푗 + 2푘) ⟹ 푖 + 푗 + 푘 퐴⃗ 30 6 6 3

퐴⃗ .퐵⃗⃗ (ii) Å⃗⃗⃗ = |퐵⃗⃗ |

|퐵⃗⃗ | = √22 + (−12) + (−3)2 ⟹ |퐵⃗⃗ | = √14

퐴⃗ . 퐵⃗⃗ = (5푖 − 푗 + 2푘). (2푖 − 푗 − 3푘) ⟹ 10 + 1 − 6 = 5

5 Å⃗⃗⃗ = √14