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Higher Maths 2 4 Circles

Higher Maths 2 4 Circles. 1. OUTCOME. SLIDE. UNIT. REMEMBER. !. Higher Maths 2 4 Circles. 2. OUTCOME. SLIDE. UNIT. NOTE. B ( x 2 , y 2 ). Distance Between Two Points. d =. √. ( x 2 – x 1 ) ². ( y 2 – y 1 ) ². +. y 2 – y 1.

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Higher Maths 2 4 Circles

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  1. Higher Maths 2 4 Circles 1 OUTCOME SLIDE UNIT

  2. REMEMBER ! Higher Maths 2 4 Circles 2 OUTCOME SLIDE UNIT NOTE B (x2,y2) Distance Between Two Points d= √ (x2–x1)² (y2–y1)² + y2 – y1 The Distance Formula Example Calculate the distance between (-2,9) and (4,-3). x2 – x1 A (x1,y1) d= √ 6² 12² + Where required, write answers as a surd in its simplest form. √ √ = = 180 6 5

  3. NOTICE ! Higher Maths 2 4 Circles 3 OUTCOME SLIDE UNIT NOTE y Points on a Circle Example Plot the following points and find a rule connecting x and y. x (5,0) (4,3) (3,4) (0,5) (-3,4) (-4,3) (-5,0) (-4,-3) (-3,-4) (0,-5) (3,-4) (4,-3) All points lie on a circle with radius 5 units and centre at the origin. For any radius... x²+y² = r² x²+y² = 25 For any point on the circle,

  4. ! LEARN THIS Higher Maths 2 4 Circles 4 OUTCOME SLIDE UNIT NOTE The Equation of a Circle with centre at the Origin For any circle with radius r and centre the origin, The ‘Origin’ is thepoint (0,0) y x x²+y² = r² origin Substitute point into equation: Example x²+y² = (-3)² +()² Show that the point (-3, ) lies on the circle with equation 7 7 = 9 + 7 = 16 x²+y² = 16 The point lies on the circle.

  5. ! LEARN THIS Higher Maths 2 4 Circles 5 OUTCOME SLIDE UNIT NOTE The Equation of a Circle with centre (a,b) y r Not all circles are centered at the origin. (a,b) For any circle with radius r and centre at the point (a,b)... x (x–a)²+ (y–b)²= r² (x–a)²+ (y–b)²= r² Example Write the equation of thecircle with centre (3,-5) and radius 2 3 . (x–3)²+ (y–(-5))²= ( )² 2 3 (x–3)²+ (y+5)²= 12

  6. ! LEARN Higher Maths 2 4 Circles 6 OUTCOME SLIDE UNIT NOTE The General Equation of a Circle Try expanding the equation of a circle with centre (-g,-f ). (x+g)2+ (y+f)2=r2 this is just a number... (x2+ 2gx + g2)+ (y2+ 2fy + f2)=r2 c=g2+f2–r2 x2+ y2+ 2gx+ 2fy+g2 + f2 – r2 = 0 r2=g2+f2–c x2+ y2 + 2gx + 2fy+ c=0 r=g2+f2–c General Equation of a Circle with center (-g,-f ) r= g2+f2–c and radius

  7. REMEMBER ! Higher Maths 2 4 Circles 7 OUTCOME SLIDE UNIT NOTE Circles and Straight Lines A line and a circle can have two, one or no points of intersection. two pointsof intersection one pointof intersection no pointsof intersection r A line which intersects a circle at only one point is at 90° to the radius and is is called a tangent.

  8. Higher Maths 2 4 Circles 8 OUTCOME SLIDE UNIT NOTE Intersection of a Line and a Circle How to find the points of intersection between a line and a circle: • rearrange the equation of the line into the form y =mx+c • substitute y =mx+c into the equation of the circle • solve the quadratic for x and substitute into mx+c to find y y = 2x Example x = 3 or -3 Substitute into y = 2x : Find the intersection of the circleand the line x2+ (2x)2 = 45 x2+ 4x2 = 45 y = 6 or -6 x2+ y2 = 45 5x2 = 45 Points of intersection are (3,6) and (-3,-6). 2x– y = 0 x2 = 9

  9. Higher Maths 2 4 Circles 9 OUTCOME SLIDE UNIT NOTE Intersection of a Line and a Circle (continued) y Example 2 Find where the line 2x– y + 8= 0 intersects the circle x2+ y2 + 4x + 2y – 20= 0 x x2+ (2x+8)2 + 4x + 2(2x+8)– 20 = 0 x2+ 4x2 + 32x + 64 + 4x + 4x+ 16 – 20= 0 5x2 + 40x+ 60= 0 x= -2 or -6 Factorise and solve Substituting into y = 2x+8 points of intersection as(-2,4) and (-6,-4). 5(x2 + 8x + 12) = 0 5(x + 2)(x + 6) = 0

  10. REMEMBER Higher Maths 2 4 Circles 10 OUTCOME SLIDE UNIT NOTE The Discriminant and Tangents -b b2–(4ac) ± x= The discriminant can be used to show that a line is a tangent: 2a • substitute into the circle equation • rearrange to form a quadratic equation • evaluate the discriminant y =mx+c b2–(4ac) Discriminant b2–(4ac)> 0 Two points of intersection b2–(4ac)= 0 r The line is a tangent b2–(4ac)< 0 No points of intersection

  11. Higher Maths 2 4 Circles 11 OUTCOME SLIDE UNIT NOTE y Circles and Tangents Example x Show that the line 3x+ y= -10 is a tangent to the circle x2+ y2 – 8x + 4y – 20= 0 x2+ (-3x–10)2 – 8x + 4(-3x–10)– 20 = 0 x2+ 9x2 + 60x +100–8x–12x– 40– 20 = 0 10x2 + 40x +40 = 0 b2– (4ac) = 402– (4 × 10 × 40) The line is a tangent to the circle since = 1600– 1600 b2– (4ac) = 0 = 0

  12. Higher Maths 2 4 Circles 12 OUTCOME SLIDE UNIT NOTE REMEMBER Equation of Tangents y–b=m(x–a) To find the equation ofa tangent to a circle: Straight Line Equation • Find the center of the circle and the point where the tangent intersects y2–y1 mradius= x2–x1 • Calculate the gradient of the radius using the gradient formula • Write down the gradient of the tangent r • Substitute the gradient of the tangent and the point of intersection into –1 mtangent= mradius y–b=m(x–a)

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