equation of a circle - wordpress.com...what is the radius of the circle represented by the equation:...
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Advanced AlgebraEquation of a Circle
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Monday, 19 October 2015 Task on Entry – Plotting Equations
Using the table and axis below, plot the graph for -
𝑥2 + 𝑦2 = 25
x -5 -4 -3 0 3 4 5
y1 4
y2 - 4
32 + 𝑦2 = 25
9 + 𝑦2 = 25
𝑦2 = 16
𝑦 = 16 𝑦 = 4
𝑦 = −4
3
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Monday, 19 October 2015 Task on Entry – Plotting Equations
x -5 -4 -3 0 3 4 5
y1
y2
00
3-3
4-4
5-5
4-4
3-3
00
𝑥2 + 𝑦2 = 25
What is the radius of the circle?
5 units
25 = 5
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Monday, 19 October 2015
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Monday, 19 October 2015
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Monday, 19 October 2015
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Equation of a Circle
Mini-whiteboards!
![Page 8: Equation of a Circle - WordPress.com...What is the radius of the circle represented by the equation: 2+ 2=40 40= 10× 4 40= 10×2 40=210 Super Challenge Hint: You need to remember](https://reader036.vdocuments.mx/reader036/viewer/2022071504/6124bf982beaa4503c6338fe/html5/thumbnails/8.jpg)
What is the radius of the circle represented by the equation:
𝑥2 + 𝑦2 = 4
4 = 2
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What is the radius of the circle represented by the equation:
𝑥2 + 𝑦2 = 100
100 =10
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What is the radius of the circle represented by the equation:
𝑥2 + 𝑦2 = 36
36 = 6
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What is the radius of the circle represented by the equation:
𝑥2 + 𝑦2 = 144
144 = 12
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What is the radius of the circle represented by the equation:
𝑥2 + 𝑦2 = 40
40 = 10 × 4
40 = 10 × 2
40 = 2 10
Super ChallengeHint: You need to remember your rules of surds
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GeometryPerpendicular Lines
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Monday, 19 October 2015 Perpendicular Lines Recap
Perpendicular lines intersect at 90 degrees
The product of the two gradients always equals - 1
× gradient of line 2 = -1gradient of line 1
gradient = 2
What is the gradient of it’s perpendicular line?
gradient = −1
2
This is called the negative reciprocal
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Write down the negative reciprocal of the following gradients
Gradient of line 1 Gradient of its perpendicular line
5
6
- 3
- 8
1
9
−1
7
−1
5
−1
6
1
3
1
8
−9
7
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GeometryCircles and Tangents
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Monday, 19 October 2015 Circles and Tangents
The diagram shows the circle 𝑥2 + 𝑦2 = 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.
(1,3)
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Monday, 19 October 2015
This is called a “normal” line.
It will intersect the tangent at 90 degrees
What is the gradient of the normal line? Hint: we know it passes through (1,3)
1
3Gradient =
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥
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:
(1,3)
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Monday, 19 October 2015
Gradient of normal = 3
If we know the gradient of the normal, we can now find out the gradient of the tangent.
3 × −1
3= -1
The tangent that intersects the circle at the
point (1,3) has a gradient of −1
3
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Monday, 19 October 2015 Circles and Tangents
Can we work out the equation of the tangent at (1,3)?
We know :
It goes through the point (1,3)
It has a gradient of −1
3
Y = m x + c
3 = 1−𝟏
𝟑× + c
Rearranging this equation we can find out that
c = 31
3
Therefore, the equation of the tangent which intersects the circle at the point (1,3) is:
𝑦 = −1
3𝑥 + 3
1
3
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Monday, 19 October 2015 Circles and Tangents
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) iny = mx + c
(1,3)
gradient = -1
3
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Independent TaskEquation of a Circle
The diagram shows the circle 𝑥2 + 𝑦2 = 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)
(3, 4)
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Monday, 19 October 2015
What is the gradient of the normal line? (we know it passes through (3,4)
Gradient = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥
Gradient of normal = 4
3
If we know the gradient of the normal, we can now find out the gradient of the tangent.
Remember:
3
4
(3, 4)
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Monday, 19 October 2015
Gradient of normal = 4
3
If we know the gradient of the normal, we can now find out the gradient of the tangent.
4
3× −
3
4= -1
The tangent that intersects the circle at the
point (3,4) has a gradient of −3
4
(3, 4)
![Page 25: Equation of a Circle - WordPress.com...What is the radius of the circle represented by the equation: 2+ 2=40 40= 10× 4 40= 10×2 40=210 Super Challenge Hint: You need to remember](https://reader036.vdocuments.mx/reader036/viewer/2022071504/6124bf982beaa4503c6338fe/html5/thumbnails/25.jpg)
Monday, 19 October 2015 Circles and Tangents
Can we work out the equation of the tangent at (3,4)?
We know :
It goes through the point (3, 4)
It has a gradient of −3
4
Y = m x + c
4 = 3−𝟑
𝟒× + c
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:
𝑦 = −3
4𝑥 + 6.25
![Page 26: Equation of a Circle - WordPress.com...What is the radius of the circle represented by the equation: 2+ 2=40 40= 10× 4 40= 10×2 40=210 Super Challenge Hint: You need to remember](https://reader036.vdocuments.mx/reader036/viewer/2022071504/6124bf982beaa4503c6338fe/html5/thumbnails/26.jpg)
GeometryCircles and Tangents
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Monday, 19 October 2015
A (2,6)
A line l is a tangent to the circle 𝑥2 + 𝑦2 = 40 at the point A.
The point A is (2,6). The tangent crosses the x-axis at P.
a) Calculate where the tangent crosses the x-axis
b) Calculate the area of OAP.
Let’s sketch this out first to help us visualise any questions we may encounter.
𝑥2 + 𝑦2 = 40
P (?, 0)
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Monday, 19 October 2015
A (2,6)a) Calculate where the tangent crosses the x-axis
2
6 Gradient of normal = 6
2= 3
Gradient of tangent = −1
3
P (?, 0)
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Monday, 19 October 2015
A (2,6)
a) Calculate where the tangent crosses the x-axisGradient of tangent = −
1
3
6 = 2−𝟏
𝟑× + c
C = 6𝟐
𝟑
Y = −𝟏
𝟑x + 6
𝟐
𝟑
Y = m x + c
The equation of our tangent is:
P (?, 0)
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Monday, 19 October 2015
A (2,6)
a) Calculate where the tangent crosses the x-axis
Y = −𝟏
𝟑x + 6
𝟐
𝟑
When the tangent crosses the x-axis, y = 0
Y = −𝟏
𝟑x + 6
𝟐
𝟑
Substitute y = 0 into tangent equation
0 = −𝟏
𝟑x + 6
𝟐
𝟑
𝒙 = 𝟐𝟎
P (20,0)
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Monday, 19 October 2015
A (2,6)
b) Calculate the area of OAP
P (𝟐𝟎, 𝟎)0
6
18
Calculate the distance AP (using Pythagoras)
= 182 + 62 = 360
2
6
Calculate the distance of OA (using Pythagoras)
= 22 + 62 = 40
0
A P
𝟒𝟎
𝟑𝟔𝟎
Area of OAP is:
360 × 40
2
= 60