Advanced Algebra. Equation of a Circle

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Bennett Simpson
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1 Advanced Algebra Equation of a Circle

2 Task on Entry Plotting Equations Using the table and axis below, plot the graph for - x 2 + y 2 = 25 x y 1 4 y y 2 = y 2 = 25 y 2 = 16 y = 16 y = 4 3 y = 4

3 Task on Entry Plotting Equations x y 1 0 y x 2 + y 2 = 25 What is the radius of the circle? 5 units 25 = 5

7 Equation of a Circle Mini-whiteboards!

8 What is the radius of the circle represented by the equation: x 2 + y 2 = 4 4 = 2

9 What is the radius of the circle represented by the equation: x 2 + y 2 = =10

10 What is the radius of the circle represented by the equation: x 2 + y 2 = = 6

11 What is the radius of the circle represented by the equation: x 2 + y 2 = = 12

12 Super Challenge Hint: You need to remember your rules of surds What is the radius of the circle represented by the equation: x 2 + y 2 = = = = 2 10

13 Geometry Perpendicular Lines

14 Perpendicular Lines Recap Perpendicular lines intersect at 90 degrees The product of the two gradients always equals - 1 This is called the negative reciprocal gradient = 1 2 gradient of line 1 gradient of line 2 = -1 gradient = 2 What is the gradient of it s perpendicular line?

15 Write down the negative reciprocal of the following gradients Gradient of line Gradient of its perpendicular line

16 Geometry Circles and Tangents

17 Circles and Tangents (1,3) The diagram shows the circle x 2 + y 2 = 10 A tangent is drawn that touches the circle at (1,3) In geometry, a tangent is a straight line that "just touches" the curve at that point.

18 This is called a normal line. 1 (1,3) 3 It will intersect the tangent at 90 degrees What is the gradient of the normal line? Hint: we know it passes through (1,3) Gradient = change in y change inx Gradient of normal = 3 1 = 3 If we know the gradient of the normal, we can now find out the gradient of the tangent. Remember:

19 Gradient of normal = 3 If we know the gradient of the normal, we can now find out the gradient of the tangent = -1 The tangent that intersects the circle at the point (1,3) has a gradient of 1 3

Circles and Tangents Y = m x + c 3 = 1 1 3 + c Rearranging this equation we can find out that Can we work out the equation of the tangent at (1,3)?

20 Circles and Tangents Y = m x + c 3 = c Rearranging this equation we can find out that Can we work out the equation of the tangent at (1,3)? We know : c = Therefore, the equation of the tangent which intersects the circle at the point (1,3) is: It goes through the point (1,3) y = 1 3 x It has a gradient of 1 3

Circles and Tangents gradient = - 1 3 (1,3) To

tangent, the gradient of the tangent and y = mx +

21 Circles and Tangents gradient = (1,3) To find the equation of the tangent: Find the gradient of the normal Find the gradient of the tangent Using the coordinates known to lie on the tangent, the gradient of the tangent and y = mx + c Find the value of c ( the y-interception) in y = mx + c

22 Independent Task Equation of a Circle (3, 4) The diagram shows the circle x 2 + y 2 = 25 A tangent is drawn that touches the circle at (3,4) Find the equation of the tangent that touches the circle at (3,4)

23 (3, 4) What is the gradient of the normal line? (we know it passes through (3,4) Gradient = change in y change inx 4 Gradient of normal = If we know the gradient of the normal, we can now find out the gradient of the tangent. Remember:

24 Gradient of normal = 4 3 (3, 4) If we know the gradient of the normal, we can now find out the gradient of the tangent = -1 The tangent that intersects the circle at the point (3,4) has a gradient of 3 4

Circles and Tangents Y = m x + c 4 = 3 3 4 + c Can we work out the equation of the

25 Therefore, the equation of the tangent which intersects the circle at the point

25 Circles and Tangents Y = m x + c 4 = c Can we work out the equation of the tangent at (3,4)? We know : Rearranging this equation we can find out that c = 6.25 Therefore, the equation of the tangent which intersects the circle at the point (3, 4) is: It goes through the point (3, 4) y = 3 4 x It has a gradient of 3 4

26 Geometry Circles and Tangents

27 x 2 + y 2 = 40 A (2,6) P (?, 0) A line l is a tangent to the circle x 2 + y 2 = 40 at the point A. The point A is (2,6). The tangent crosses the x-axis at P. Let s sketch this out first to help us visualise any questions we may encounter. a) Calculate where the tangent crosses the x-axis b) Calculate the area of OAP.

28 A (2,6) a) Calculate where the tangent crosses the x-axis 2 6 Gradient of normal = 6 2 = 3 P (?, 0) Gradient of tangent = 1 3

29 a) Calculate where the tangent crosses the x-axis Gradient of tangent = 1 3 Y = m x + c A (2,6) 6 = c P (?, 0) C = The equation of our tangent is: Y = 1 3 x

30 Y = 1 3 x a) Calculate where the tangent crosses the x-axis A (2,6) When the tangent crosses the x-axis, y = 0 Y = 1 3 x P (20,0) Substitute y = 0 into tangent equation 0 = 1 3 x x = 20

b) Calculate the area of OAP Calculate the distance AP (using Pythagoras) A (2,6) 18 = 18 2 + 6 2 = 360 6 6

31 b) Calculate the area of OAP Calculate the distance AP (using Pythagoras) A (2,6) 18 = = Calculate the distance of OA (using Pythagoras) = = P (20, 0) 0 Area of OAP is: A 360 P = 60

10-2 Circles. Warm Up Lesson Presentation Lesson Quiz. Holt Algebra2 2

10-2 Circles. Warm Up Lesson Presentation Lesson Quiz. Holt Algebra2 2 10-2 Circles Warm Up Lesson Presentation Lesson Quiz Holt Algebra2 2 Warm Up Find the slope of the line that connects each pair of points. 1. (5, 7) and ( 1, 6) 1 6 2. (3, 4) and ( 4, 3) 1 Warm Up Find