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Advanced AlgebraEquation of a Circle
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
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
Monday, 19 October 2015
Monday, 19 October 2015
Monday, 19 October 2015
Equation of a Circle
Mini-whiteboards!
What is the radius of the circle represented by the equation:
𝑥2 + 𝑦2 = 4
4 = 2
What is the radius of the circle represented by the equation:
𝑥2 + 𝑦2 = 100
100 =10
What is the radius of the circle represented by the equation:
𝑥2 + 𝑦2 = 36
36 = 6
What is the radius of the circle represented by the equation:
𝑥2 + 𝑦2 = 144
144 = 12
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
GeometryPerpendicular Lines
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
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
GeometryCircles and Tangents
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)
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)
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
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
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
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)
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)
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)
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
GeometryCircles and Tangents
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)
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)
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)
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)
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
