A 2 Lines and Circles I Lines A

A 2 : Lines and Circles I (Lines) A 2. 1 Lines Parallel to coordinate axes: • Horizontal line–parallel to x-axisy is given by y=b, where(0, b) is y-intercept. y=0 gives x-axis (0, b) • Vertical line-parallel to y-axis is given by x=a, where (a, 0) is x-intercept. x=0 gives y-axis. x=a y=b (a, 0) Fig 1 x

• Non Vertical Lines : Take A(x 1, y 1) , B(x 2, y 2) on the line. Let P(x, y) be a general point on the line. From Fig 2 ∆ACB and ∆BQP are similar. P(x, y) y A(x 1, y 1) B(x 2, y 2) Run Q(x, y 2) Rise C(x 2, y 1) x Fig 2 : . CB/AC=QP/BQ= Rise/Run = Vertical. Change/Horizontal Change = const (1)Characterizes straight lines, and the constant is called the gradient or slope of the straight line. it is independent of the choice of A& B.

• If Run≠ 0 then Gradient = Rise/Run =Vertical Change/Horizontal Change (2) • If Run=0 (No horizontal Change) we get a vertical line with infinite slope. • If Rise=0 (No vertical change) we get a horizontal line with slope=0.

• A 2. 2 Various Equations of lines: (i) With slope=m and passing through. A(x 1’y 1): m=(Rise from A to P)/(Run from A to P)= (y P(x, y) -y 1)/(x-x 1) (y-y 1)=m(x-x 1) Rise Or y= y 1+ m(x-x 1) (3) A(x 1, y 1) (x, y 1) Run Or y=mx+(y 1 -mx 1) Fig 3

(ii) With gradient m and y- intercept (0, c) A particular case of (i) With x 1=0 , y 1=c y=mx+c : . (3) gives: y= m x+ c (4) (iii) Passing through A(x 1, y 1), B (x 2, y 2) (0, c) y-intercept Here m=[ y 2 -y 1/ x 2 -x 1]) Rewrite (3) as 0 x-intercept y-y 1 = [(y 2 -y 1)/x 2 -x 1)]=(x-x 1) (5) [Do Example(1. 1) and activities (1. 1)) ـــــــ 1. 4)] • (iv) Note a straight line is given by a linear equation A x+By+C=0 (6) And conversely. (slope=-A/B) y-intercept(0, -C/B), x-intercept (-C/A, 0)

A 2. 3 Relations between lines: If ℓ 1 has slope m 1 and ℓ 2 has slope m 2: • If m 1= m 2 then ℓ 1 //ℓ 2: and if ℓ 1∩ℓ 2=ø then ℓ 1 and ℓ 2 don’t meet and if ℓ 1∩ℓ 2≠ ø then ℓ 1Ξℓ 2 • If m 1≠m 2 then ℓ 1 and ℓ 2 intersect in one point only. • Perpendicular lines: ℓ 1 with slope m, and ℓ 2 with slope ℓ 2 are perpendicular if and only if m 1 m 2= - 1, or one line is horizontal (with slope = 0) and the other is vertical( with Infinite slope). [Do activities 1. 5 , 1. 6 ]

• Example: For the given pairs of lines find points of intersection , if any. a) ℓ 1: 3 x+2 y=3, ℓ 2: 6 x+4 y=5 b) ℓ 1: 3 x+2 y=3, ℓ 3: 6 x+4 y=6 c) ℓ 1: 3 x+2 y=3, ℓ 4: 2 x-3 y=-11 Solutions : In a) & b) m 1=-3/2, m 2=m 3=-6/4=m 1. So lines ℓ 1 , ℓ 2 and ℓ 3 are parallel. But in a) 3 x+2 y=3, 6 x+4 y=5 has no solution. So ℓ 1 does’nt intersect ℓ 2. Whereas in b) 3 x+2 y=3 is the same equation as 6 x+4 y=6 So ℓ 1Ξℓ 2. In c) m 1= -3/2, m 4=2/3 : . m 1 m 4=-1 and ℓ 1┴ℓ 4. Solving the equations we get x=-1, y=3. Therefore ℓ 1 and ℓ 4 intersect at (1, 3). [Do activities (1. 7) , (1. 8)].

II CIRCLES B(x 2, y 2) y • A 2. 4; Distance formula: d By Pythagoras’ formula (x , y ) A(x , y ) The distance d between x any two points A(x 1, y 1) Fig 5 and B(x 2, y 2) is given by d 2= (x 2 -x 1)2 +(y 2 -y 1)2 (7) • A circle with centre C(a, b) and radius r is given by (x-a)2 +(y-b)2 =r 2 (8) [Do activities 2. 3 , 2. 4 ] 1 1 2 1

• Midpoint Rule: The midpoint of the segment between A(x 1, y 1) and B(x 2, y 2) has coordinates (1/2(x 1+x 2), 1/2(y 1+y 2)). • To find the equation of the circle determined by 3 noncollinear points A, B, C: (i)Find midpoints of AB, BC. (ii) Find equations of the lines perpendicular to each segment through its mid point (iii) The point of intersection of these lines gives the centre of the circle. (iv) The distance between the centre and either point A, B or C gives the radius. [ Do example 2. 1 and activity 2. 5].

A 2. 5 General Equation of circle : Expanding brackets in (8) : (x-a)2 + (y-b)2 = r 2 gives an equation of the form x 2 +y 2 +Ax +By +C = 0 (9) a second degree equation in x and y with no mixed term xy. Conversely (9) represents a circle (or an empty set ). This could be shown by completing the square. A 2. 6 Completing the square : For x 2 +2 px we write: x 2 +2 px = (x +p)2 –p 2 (10)

• Ex: Verify that the equation : x 2 – 4 x + y 2 + 2 y + 4 =0 represents a circle, find its centre and radius. Solution : complete the square in x and y : x 2 – 4 x= (x-2)2 -4; y 2 +2 y =(y+1)2 -1 : . x 2 -4 x +y 2 +2 y +4 =(x -2)2 -4 +(y +1)2 -1+4=0 (x -2)2 + (y +1)2 =1 Equation of a circle with radius =1 and centre C(2, -1). [ Do activity 2. 6 ].

A 2. 7 Intersections of circles and lines. A given circle intersects a line y ℓ 1 in either : ℓ 2 (1) Two points (e. g. ℓ 1) ℓ 3 x (2) one point (tangent)e. g. ℓ 2 Fig 6 (3) No point (e. g. ℓ 3) • To find the points of intersection: (i) From equation of line solve for one variable (ii) Substitute in equation of circle. (iii) Solve the resulting quadratic equation. [ Do example 2. 3 and activity 2. 7].

• Two given circles intersect either in : (i) Two points (a) or a (ii) one point (b) or b (iii) no point (c) c Fig 7 To find points of intersection solve the two equations simultaneously to get a linear equation then substitute in either circle as above.

III Trigonometry A 2. 8 Sine and cosine : Let S 1 = { (x, y): x 2+y 2=1} be the unit circle with centre at O and radius 1. P(x, y)Є S 1 y P(x, y) Let θ be the angle between (o, y) positive x-axis and OP, θ (x, o) 0 measured anticlockwise. Define: Fig 8 cos θ=x, sin θ=y (11) Θ is measured in radians or in degrees. 2π radians = 360 degrees (12). x

From definition it follows that , for all θ: cos 2 θ +sri 2 θ =1 (13) cos (θ +2 n π) =cos θ , n any integer (14) sin (θ +2 nπ) =sin θ , n any integer (15) -1 ≤ sin θ≤ 1, -1≤ cos θ ≤ 1 (16) y Also: cos θ<o Sin θ>o cos θ>o Sin θ>o x cos θ<o Sin θ<o cos θ>o Sin θ<o Fig 9

A 2. 9 y Reflections: y y 1 o θ -θ x Q=(y, x) P(x. y) Q=(-x, y) P(x. y) 1 θ X=y P(x. y) θ x o π -θ Q=(-x, y) Fig 10 shows reflections of the points P(x, y) on the circle about: A. x –axis to give Q (x, y). B. y –axis to give Q (- x y). C. The line y=x to give Q (y x). x ½ π -θ

Using these symmetries we can deduce that for all θ: cos(-θ)=cosθ(even), sin(-θ)=-sin θ(odd) (17) cos (π-θ)= -cos θ, sin(π-θ)= sin θ (18) cos (π/2 -θ)= sinθ, sin (π/2 -θ)= cos θ (19) [Do activity 3. 2 ]

A 2. 10 Certain Values of sinθ and cosθ : Given in Table : θ 0 π/6 π/4 π/3 π/2 π 3π/2 2π cosθ 1 √ 3/2 √ 2/2 1/2 0 -1 0 1 sinθ 0 1 0 -1 0 1/2 √ 2/2 √ 3/2 A 2. 11 Rest of Trigonometric functions : tan θ= sinθ/cosθ, cotθ= cosθ/sinθ= 1/tanθ sec θ=1/cosθ, cosecθ=1/sinθ Their values and properties could be deduced from those of cos θ and sin θ.

A 2. 12 Calculations with Triangles. y A P(x, y) y=sinθ 1 θ 0 X=COSθ X B x Fig 11 Shows part of the unit circle with a point P(x, y) on it and a right angled triangle. Triangles OPX and OAB are similar. Hence OB/OA=OX/OP=cosθ, BA/OA=XP/OP= sinθ.

In Δ 0 AB : OA is called hypotenuse (hyp) OB is called adjacent (adj) AB is called opposite (opp) So we have in a right angled triangle : sinθ = opp/hyp ; cosθ = adj/hyp tanθ = opp/adj , cotθ = adj/opp secθ= hyp/adj , cosecθ=hyp/opp. [Do activity 3. 5].

Parametric equations: A 2. 13 parametric equations of lines: x and y are given as functions of a third variable t , called the parameter. a) For a line with slope m and passing thru (x 1, y 1)a suitable parameterization is: x=t+x 1 , y=mt+y 1. (20) Here (x, y)=(x 1, y 1) when t=o and t gives the run from (x 1, y 1) to (x, y)

b) For a line through (x 1, y 1) and (x 2, y 2) a suitable parameterization is: x=x 1+t (x 2 -x 1), y=y 1+t(y 2 -y 1) (21) When t=o (x, y)=(x 1, y 1) y and when t=1 (x, y)=(x 2, y 2) . t . A t=0 . B (x 2, y 2) (x 1, y 1) (x, y) 0 [Do activity 4. 1] *Eliminating the parameter t results in a linear equation in x and y [Do example 4, 1 and activity 4, 2] x

A 2. 4 Parameteic equations of circles For a circle of radius r and ceutre C (a, b) a suitable parameterization is given by: x=a+r cos (kt) o≤kt≤ 2π y=b+r sin (kt) o≤t≤ 2π/k (22) Restricting range of t gives portions of a circle. (22) Could be written as x=a+r cos θ , y=b+r sin θ , where θ=kt [Do activities 4. 4 & 4. 5]