Tangents + intersections
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- The document discusses finding the equation of a tangent line to a circle given a point of tangency, as well as finding points of intersection between a line and a circle. - The key steps are to find the center and radius of the circle from its equation, then use properties of tangents (gradient of radius = -1/gradient of tangent) to determine the gradient of the tangent line. - To find intersections, set the line and circle equations equal and solve using substitution or factorizing, looking for real number solutions. If only one solution is found, the line is tangent to the circle.
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The document lists various geometric shapes including triangle, rectangle, circle, cylinder, cone, ring, hexagon, octagram, trapezium, and also mentions a presentation was made by several students and teachers on the topic of shapes.
Part 3 Andrew Gonzales
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Descriptive Geometry and Drawing
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The document defines key terms related to circles: a circle consists of all points equidistant from a center point and is named after its center; the radius is the distance from the center to the circle; a chord connects two points on the circle; a diameter passes through the center; a secant contains a chord; a tangent intersects at only one point; congruent circles have the same radii; and concentric circles share the same center. Circles can be inscribed inside or circumscribed about polygons based on whether vertices or sides are tangent to the circle.
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35182797 additional-mathematics-form-4-and-5-notes
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1 of 402.5 PointsUse Cramer’s Rule to solve the following syst.docx
1 of 40
2.5 Points
Use Cramer’s Rule to solve the following system.
x + 2y = 3
3x - 4y = 4
A. {(3, 1/5)}
B. {(5, 1/3)}
C. {(1, 1/2)}
D. {(2, 1/2)}
2 of 40
2.5 Points
Solve the following system of equations using matrices. Use Gaussian elimination with back substitution or Gauss-Jordan elimination.
x + y - z = -2
2x - y + z = 5
-x + 2y + 2z = 1
A. {(0, -1, -2)}
B. {(2, 0, 2)}
C. {(1, -1, 2)}
D. {(4, -1, 3)}
3 of 40
2.5 Points
Use Cramer’s Rule to solve the following system.
2x = 3y + 2
5x = 51 - 4y
A. {(8, 2)}
B. {(3, -4)}
C. {(2, 5)}
D. {(7, 4)}
4 of 40
2.5 Points
Use Cramer’s Rule to solve the following system.
4x - 5y = 17
2x + 3y = 3
A. {(3, -1)}
B. {(2, -1)}
C. {(3, -7)}
D. {(2, 0)}
5 of 40
2.5 Points
Use Cramer’s Rule to solve the following system.
4x - 5y - 6z = -1
x - 2y - 5z = -12
2x - y = 7
A. {(2, -3, 4)}
B. {(5, -7, 4)}
C. {(3, -3, 3)}
D. {(1, -3, 5)}
6 of 40
2.5 Points
Use Cramer’s Rule to solve the following system.
3x - 4y = 4
2x + 2y = 12
A. {(3, 1)}
B. {(4, 2)}
C. {(5, 1)}
D. {(2, 1)}
Reset Selection
7 of 40
2.5 Points
Use Cramer’s Rule to solve the following system.
x + y + z = 0
2x - y + z = -1
-x + 3y - z = -8
A. {(-1, -3, 7)}
B. {(-6, -2, 4)}
C. {(-5, -2, 7)}
D. {(-4, -1, 7)}
8 of 40
2.5 Points
Solve the following system of equations using matrices. Use Gaussian elimination with back substitution or Gauss-Jordan elimination.
3x1 + 5x2 - 8x3 + 5x4 = -8
x1 + 2x2 - 3x3 + x4 = -7
2x1 + 3x2 - 7x3 + 3x4 = -11
4x1 + 8x2 - 10x3+ 7x4 = -10
A. {(1, -5, 3, 4)}
B. {(2, -1, 3, 5)}
C. {(1, 2, 3, 3)}
D. {(2, -2, 3, 4)}
9 of 40
2.5 Points
Solve the following system of equations using matrices. Use Gaussian elimination with back substitution or Gauss-Jordan elimination.
x + y + z = 4
x - y - z = 0
x - y + z = 2
A. {(3, 1, 0)}
B. {(2, 1, 1)}
C. {(4, 2, 1)}
D. {(2, 1, 0)}
10 of 40
2.5 Points
Solve the system using the inverse that is given for the coefficient matrix.
2x + 6y + 6z = 8
2x + 7y + 6z =10
2x + 7y + 7z = 9
The inverse of:
2
2
2
6
7
7
6
6
7
is
7/2
-1
0
0
1
-1
-3
0
1
A. {(1, 2, -1)}
B. {(2, 1, -1)}
C. {(1, 2, 0)}
D. {(1, 3, -1)}
Reset Selection
11 of 40
2.5 Points
Use Gaussian elimination to find the complete solution to the following system of equations, or show that none exists.
2w + x - y = 3
w - 3x + 2y = -4
3w + x - 3y + z = 1
w + 2x - 4y - z = -2
A. {(1, 3, 2, 1)}
B. {(1, 4, 3, -1)}
C. {(1, 5, 1, 1)}
D. {(-1, 2, -2, 1)}
12 of 40
2.5 Points
Use Cramer’s Rule to solve the following system.
x + y = 7
x - y = 3
A. {(7, 2)}
B. {(8, -2)}
C. {(5, 2)}
D. {(9, 3)}
13 of 40
2.5 Points
Use Gaussian elimination to find the complete solution to each system.
x1 + 4x2 + 3x3 - 6x4 = 5
x1 + 3x2 + x3 - 4x4 = 3
2x1 + 8x2 + 7x3 - 5x4 = 11
2x1 + 5x2 - 6x4 = 4
A. {(-47t + 4, 12t, 7t + 1, t)}
B. {(-37t + 2, 16t, -7t + 1, t)}
...
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Alg2 lesson 8-6
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1) A parabola with vertex (1,4) and focus (3,4)
2) A circle with center (0,-3) and radius 6
3) An ellipse with center (1,0), vertices of (1±2/√8,0), and foci of (3,0) and (-1,0)
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This document provides formulas and examples for calculating the distance and midpoint between two points.
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Jawapan matriks spM LATIH TUBI 2015
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BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
Accounting for Restricted Grants When and How To Record Properly
Accounting for Restricted Grants When and How To Record Properly
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A Free 200-Page eBook ~ Brain and Mind Exercise.pptx
CapTechTalks Webinar Slides June 2024 Donovan Wright.pptx
CapTechTalks Webinar Slides June 2024 Donovan Wright.pptx
Tangents + intersections
- 1. Block 1 Circles and Tangents
- 2. What is to be learned? • How to find the equation of a tangent to a circle
- 3. Some revision m1m2 = -1 For equation need m (a , b) use y – b = m(x – a)
- 4. x2 + y2 + 2x – 4y – 5 = 0 Equation of Tangent at (2 , 1) get centre x2 + y2 + 2gx + 2fy + c = 0 2g = 2 , 2f = -4 g = 1 , f = -2 Centre (-1 , 2)
- 5. Equation of Tangent at (2 , 1) Centre (-1 , 2) (2 , 1) (-1 , 2) Need (a , b) m m = 1 – 2 2 – (-1) = -1 /3 so m of tangent = 3 use y – b = m(x – a)
- 6. x2 + y2 – 6x – 14y – 3 = 0 Equation of Tangent at (5 , 8) get centre x2 + y2 + 2gx + 2fy + c = 0 2g = -6 , 2f = -14 g = -3 , f = -7 Centre (3 , 7)
- 7. Equation of Tangent at (5 , 8) Centre (3 , 7) (5 , 8) (3 , 7) Need (a , b) m m = 7 – 8 3 – 5 = ½ so m of tangent = -2
- 8. y – b = m(x – a) m = -2, (a , b) = (5 , 8) y – 8 = -2(x – 5) y – 8 = -2x + 10 y = -2x + 18
- 9. Equation of a Tangent to a Circle If given circle equation and point where tangent meets circle Tactics • From equation find centre • Find gradient of radius • Use this to get gradient of tangent • Get equation of tangent (m1m2 = -1) y – b = m(x – a)
- 10. x2 + y2 + 2x – 6y + 8 = 0 Equation of Tangent at (1 , 2) get centre x2 + y2 + 2gx + 2fy + c = 0 2g = 2 , 2f = -6 g = 1 , f = -3 Centre (-1 , 3)
- 11. Equation of Tangent at (1 , 2) Centre (-1 , 3) (1 , 2) (-1 , 3) Need (a , b) m (1 , 2) m = 2 – 3 1 – (-1) = -1 /2 so m of tangent = 2 2 Now use y – b = m(x – a)
- 12. x2 + y2 + 12x – 10y – 3 = 0 Find equation of Tangent at (-2 , 4) get centrex2 + y2 + 2gx + 2fy + c = 0 2g = 12 , 2f = -10 g = 6 , f = -5 Centre (-6 , 5) Key Question
- 13. Equation of Tangent at (-2 , 4) Centre (-6 , 5) (-2 , 4) (-6 , 5) Need (a , b) m m = 5 – 4 -6 + 2 = -¼ so m of tangent = 4
- 14. y – b = m(x – a) m = 4, (a , b) = (-2 , 4) y – 4 = 4(x + 2) y – 4 = 4x + 8 y = 4x + 12
- 15. Intersections of Lines and Circles
- 16. What is to be learned? • How to find the point(s) of intersection between line and circle If any!
- 17. Point of Intersection x2 + y2 = 5 y = 2x + 5 simultaneous equations y = y substitution
- 18. Point of Intersection x2 + y2 = 5 y = 2x + 5
- 19. Point of Intersection x2 + y2 = 5 y = 2x + 5 x2 + (2x + 5)2 = 5 (2x + 5)2 =(2x + 5)(2x + 5) =4x2 + 20x + 25 = x2 + 4x2 + 20x +25 = 5 = 5x2 + 20x +20 = 0 = 5(x2 + 4x + 4) = 0 = 5(x + 2)(x + 2) = 0 x = -2 y? y= 2(-2) + 5 = 1 (-2 , 1) One Point of Intersection Tangent No solution if b2 – 4ac < 0
- 20. Intersection of Line and Circle • Must use substitution • If only one point of intersection Tangent • If no points of intersection b2 – 4ac < 0
- 21. x2 + y2 + 2x – 2y – 11 = 0 5y = x – 7 x = 5y + 7 sub for x
- 22. x2 + y2 + 2x – 2y – 11 = 0 5y = x – 7 x = 5y + 7 (5y + 7)2 + y2 + 2(5y + 7) – 2y – 11 = 0 25y2 + 70y + 49 + y2 + 10y + 14 – 2y – 11 = 0 26y2 + 78y + 52 = 0 26(y2 + 3y + 2) = 0 26(y + 2)(y + 1) = 0 y = -2 or -1 y = -2 x = 5(-2) + 7 y = -1 x = 5(-1) + 7 (-3 , -2) (2 , -1) * *(5y + 7)(5y + 7) using sub for x
- 23. Find Point of Intersection, and state what type of line it is in relation to the circle. x2 + y2 = 18 y = x + 6 x2 + (x + 6)2 = 18 = x2 + x2 + 12x +36 = 18 = 2x2 + 12x +18 = 0 = 2(x2 + 6x + 9) = 0 = 2(x + 3)(x + 3) = 0 x = -3 y? y = -3 + 6 = 3 (-3 , 3) One Point of Intersection Tangent Key Question