MAT 104
ELEMENTARY MATHEMATICS III
3 CREDITS
CO-ORDINATE GEOMETRY (CONT’D)
CIRCLES
The equation of a circle with centre (𝑎, 𝑏) and radius 𝑟 is given by (𝑥 − 𝑎)2 +
(𝑦 − 𝑏)2 = 𝑟 2
When the circle has its centre at origin, the equation of the circle becomes or
reduces to
𝑥2 + 𝑦2 = 𝑟2
Because, circle with centre at origin means the coordinates of the centre is (0, 0) ⟹
(𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟 2
(𝑥 − 0)2 + (𝑦 − 0)2 = 𝑟 2
𝑥2 + 𝑦2 = 𝑟2
Examples
1. Find the equation of the circle which has its center at origin and a radius of 5.
Solution
Simply from 𝑥 2 + 𝑦 2 = 𝑟 2 ⟹ 𝑥 2 + 𝑦 2 = 52
Thus 𝑥 2 + 𝑦 2 = 25
2. Find the equation of the circle with center (2, 5) and a radius of 3.
Solution
From (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟 2 ⟹ (𝑥 − 2)2 + (𝑦 − 5)2 = 32
⟹ 𝑥 2 + 𝑦 2 − 4𝑥 − 10𝑦 + 20 = 0
3. Find the equation of a circle with centre (−2, 3) which passes through the
point (1, 2)
Solution
Length of radius, 𝑟 = √(−2 − 1)2 + (3 − 2)2
𝑟 = √−32 + 12
𝑟 = √10
𝑟 2 = 10
Using (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 with the circle centre (−2, 3) and radius 𝑟 = √10
The equation of the circle is (𝑥 + 2)2 + (𝑦 − 3)2 = 10
⟹ 𝑥 2 + 𝑦 2 + 4𝑥 − 6𝑦 + 3 = 0
4. Find the radius and centre of the circle whose equation is given by
3𝑥 2 + 3𝑦 2 − 24𝑥 + 12𝑦 + 11 = 0
Solution
𝑥 2 + 𝑦 2 − 8𝑥 + 4𝑦 +
11
3
=0
𝑥 2 + 𝑦 2 − 8𝑥 + 4𝑦 = −
11
3
Completing squares
𝑥 2 − 8𝑥 + 16 + 𝑦 2 + 4𝑦 + 4 = −
(𝑥 − 4)2 + (𝑦 + 2)2 =
11
3
+ 16 + 4
49
3
Comparing with (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟 2 ⟹ 𝑎 = 4, 𝑏 = −2, 𝑟 = √
Circle centre (4, −2) and radius
7
√3
49
3
EQUATION OF A CIRCLE GIVEN INFORMATION ABOUT THE DIAMETER
The equation of a circle can also be deduced if the information about the points
which the diameter passes through is known. In summary the equation of a circle
whose diameter is the join of (𝑥1 , 𝑥2 ) and (𝑦1 , 𝑦2 ) is given by the equation
(𝑥 − 𝑥1 )(𝑥 − 𝑥2 ) + (𝑦 − 𝑦1 )(𝑦 − 𝑦2 ) = 0
Examples
1. Find the equation of a circle drawn on diameter 𝐴𝐵 where 𝐴 is (−1, 3) and 𝐵
is (3, 2)
Solution
5
Mid-point of 𝐴𝐵; (1, ) ⟹ centre of circle
2
5
Thus, 𝑟 2 = (1 + 1)2 + ( − 3)2
2
1
𝑟 2 = 22 + (− )2
2
𝑟2 =
17
4
(Note that to get the length of 𝑟, you can use any of the two points with the
5
5
2
2
circle centre i.e. {(−1, 3) and (1, )} or {(3, 2) and (1, )}
5
17
2
4
Equation of circle (𝑥 − 1)2 + (𝑦 − )2 =
𝑥 2 − 2𝑥 + 1 + 𝑦 2 − 5𝑦 +
25
4
=
17
4
𝑥 2 + 𝑦 2 − 2𝑥 − 5𝑦 + 3 = 0
Using the formula method
(𝑥 − 𝑥1 )(𝑥 − 𝑥2 ) + (𝑦 − 𝑦1 )(𝑦 − 𝑦2 ) = 0
⟹ (𝑥 + 1)(𝑥 − 3) + (𝑦 − 3)(𝑦 − 2) = 0
⟹ (𝑥 2 − 2𝑥 − 3) + (𝑦 2 − 5𝑦 + 6) = 0
⟹ 𝑥 2 + 𝑦 2 − 2𝑥 − 5𝑦 + 3 = 0
2. The equation of a circle is 𝑥 2 + 𝑦 2 − 6𝑥 + 8𝑦 = 24. If the circle which passes
through the point (−1, 1), find the coordinates of the other end of the
diameter and the equation of the diameter.
Solution
𝑥 2 + 𝑦 2 − 6𝑥 + 8𝑦 = 24
𝑥 2 − 6𝑥 + 9 + 𝑦 2 + 8𝑦 + 16 = 24 + 9 + 16
(𝑥 − 3)2 + (𝑦 + 4)2 = 49
Thus the circle centre is (3, −4) ⟹ 𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
Note that the first end of the diameter is (−1, 1), so the coordinates of the
other end of the diameter is obtained as
3=
3=
1
2
1
2
(𝑥1 + 𝑥2 )
−4 =
(−1 + 𝑥2 )
−4 =
𝑥2 = 7
1
2
(𝑦1 + 𝑦2 )
1
2
(1 + 𝑦2 )
𝑦2 = −9
Thus the coordinates of the other end of the diameter is (7, −9).
Now since the coordinates of the two points of the diameter is known (−1, 1) and
(7, −9). the equation of the line (i.e. the diameter) is obtained using the two-point
form
𝑦 − 𝑦1 =
𝑦−1=
𝑦2 − 𝑦1
𝑥2 − 𝑥1
−9 −1
7+ 1
5
(𝑥 − 𝑥1 )
(𝑥 + 1)
𝑦 − 1 = − (𝑥 + 1) ⟹ 4𝑦 + 5𝑥 + 9 = 0
4
Alternatively,
To find the equation of the diameter, we can use just the coordinates of the point
that we were given i.e. (−1, 1) and the coordinates of the circle centre i.e. (3, −4),
which is equally the mid-point of the diameter.
Using the two-point form,
𝑦 − 𝑦1 =
𝑦−1=
𝑦2 − 𝑦1
𝑥2 − 𝑥1
−4 −1
3+1
(𝑥 − 𝑥1 )
(𝑥 + 1)
5
𝑦 − 1 = − (𝑥 + 1) ⟹ 4𝑦 + 5𝑥 + 9 = 0
4
(The reason for this is because the three points – the two ends of the diameter and
the circle centre i.e. (−1, 1), (7, −9) and (3, −4) lie on the same straight line and as
such are collinear points meaning they will always have the same gradient).
GENERAL FORM OF THE EQUATION OF A CIRCLE
The equation of a circle with centre (𝑎, 𝑏) and radius 𝑟 is (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 =
𝑟 2 which gives
𝑥 2 + 𝑦 2 − 2𝑎𝑥 − 2𝑏𝑦 + 𝑎2 + 𝑏 2 − 𝑟 2 = 0
written in the form 𝑥 2 + 𝑦 2 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐 = 0, which is the general standard
form of the equation of a circle. Any of 𝑔, 𝑓 and 𝑐 can be 0
With This general form of the equation of a circle, the equation of a circle can be
determined given that it passes through three points by substituting the (𝑥, 𝑦)
values of each point into the equation above and solving the three equations
simultaneously.
Examples
1. Find the equation of a circle through the points (6, 1), (3,2) and (2, 3)
Using 𝑥 2 + 𝑦 2 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐 = 0
For the point (6, 1), we have 62 + 12 + 2(6)𝑔 + 2(1)𝑓 + 𝑐 = 0
(𝑖)
For the point (3, 2), we have 32 + 22 + 2(3)𝑔 + 2(2)𝑓 + 𝑐 = 0
(𝑖𝑖)
For the point (2, 3), we have 22 + 32 + 2(2)𝑔 + 2(3)𝑓 + 𝑐 = 0
(𝑖𝑖𝑖)
This becomes
37 + 12𝑔 + 2𝑓 + 𝑐 = 0
(𝑖𝑣)
13 + 6𝑔 + 4𝑓 + 𝑐 = 0
(𝑣)
13 + 4𝑔 + 6𝑓 + 𝑐 = 0
(𝑣𝑖)
Subtract equation (𝑣𝑖) from (𝑣), we have
2𝑔 − 2𝑓 = 0 ⟹ 𝑔 = 𝑓
(𝑣𝑖𝑖)
Subtract equation (𝑣) from (𝑖𝑣), we have
24 + 6𝑔 − 2𝑓 = 0
(𝑣𝑖𝑖𝑖)
Substitute for 𝑔 in equation (𝑣𝑖𝑖𝑖)
24 + 6𝑓 − 2𝑓 = 0 (since 𝑔 = 𝑓)
4𝑓 = −24
𝑓 = −6 ⟹ 𝑔 = −6
Solving for 𝑐, we have 𝑐 = 47
Thus the equation of the circle is 𝑥 2 + 𝑦 2 − 12𝑥 − 12𝑦 + 47 = 0
2. Find the equation of a circle through the points (0, 0), (3, 1) and (5, 5)
Using 𝑥 2 + 𝑦 2 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐 = 0
For the point (0, 0), we have 02 + 02 + 2(0)𝑔 + 2(0)𝑓 + 𝑐 = 0 ⟹ 𝑐 = 0
(𝑖)
For the point (3, 1), we have 32 + 12 + 2(3)𝑔 + 2(1)𝑓 + 𝑐 = 0
⟹ 10 + 6𝑔 + 2𝑓 + 𝑐 =
0
(𝑖𝑖)
For the point (5, 5), we have 52 + 52 + 2(5)𝑔 + 2(5)𝑓 + 𝑐 = 0
⟹ 50 + 10𝑔 + 10𝑓 + 𝑐 =
0
(𝑖𝑖𝑖)
Solve equations (𝑖𝑖) and (𝑖𝑖𝑖) simultaneously, we have 𝑔 = 0 and 𝑓 = −5
Thus the equation of the circle is 𝑥 2 + 𝑦 2 − 10𝑦 = 0
CONCYCLIC POINTS
Four or more points are said to be concyclic if they lie on the same circle.
Example
Determine if the points (5, 2), (2, 3), (−3, −2) and (6, −5) are concyclic.
Solution
Using 𝑥 2 + 𝑦 2 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐 = 0
For the point 52 + 22 + 2(5)𝑔 + 2(2)𝑓 + 𝑐 = 0
(𝑖)
For the point 22 + 32 + 2(2)𝑔 + 2(3)𝑓 + 𝑐 = 0
(𝑖𝑖)
For the point −32 + −22 + 2(−3)𝑔 + 2(−2)𝑓 + 𝑐 = 0
(𝑖𝑖𝑖)
Solving equations (𝑖), (𝑖𝑖) and (𝑖𝑖𝑖) simultaneously, 𝑔 = −2, 𝑓 = 2, 𝑐 = −17
Thus the equation of the circle is 𝑥 2 + 𝑦 2 − 4𝑥 + 4𝑦 − 17 = 0
Now if the points are concyclic input the coordinates of the fourth point into the
equation of the circle to verify if this fourth point satisfies the equation
Therefore, we put the point (6, −5) into the equation 𝑥 2 + 𝑦 2 − 4𝑥 + 4𝑦 − 17 = 0
We have, 62 + (−5)2 − 4(6) + 4(−5) − 17 = 0
⟹ 36 + 25 − 24 − 20 − 17 = 0
This shows that the points are concyclic.
POINTS INSIDE OR OUTSIDE A CIRCLE
Consider the function 𝑓(𝑥, ) = (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 − 𝑟 2 .
𝑓(𝑥, 𝑦) = 0 defines a circle with radius 𝑟 and centre (𝑎, 𝑏). If the point (ℎ, 𝑘) lies
inside the circle then
(ℎ − 𝑎)2 + (𝑘 − 𝑏)2 < 𝑟 2 i. e. (ℎ − 𝑎)2 + (𝑘 − 𝑏)2 − 𝑟 2 < 0 or 𝑓(ℎ, 𝑘) < 0.
So if (ℎ, 𝑘) lies inside the circle given by 𝑓(𝑥, 𝑦) = 0, then 𝑓(ℎ, 𝑘) < 0 and if the
point (ℎ, 𝑘) lies outside the circle then 𝑓(ℎ, 𝑘) > 0.
Example
1. Are the points 𝐴 is (−1, −1) and 𝐵 is (5, 2) inside or outside the circle
𝑥 2 + 𝑦 2 − 3𝑥 + 4𝑦 = 12 .
Let 𝑓(𝑥, 𝑦) = 𝑥 2 + 𝑦 2 − 3𝑥 + 4𝑦 − 12
𝑓(1, −1) = 1 + 1 − 3 − 4 − 12
= −17 < 0 ⟹ 𝐴 is inside the circle
𝑓(5, 2) = 25 + 4 − 15 + 8 − 12
= 10 > 0 ⟹ 𝐵 is outside the circle
POINT OF INTERSECTION OF A STRAIGHT LINE AND A CIRCLE
The coordinates of the point of intersection of a straight line and a circle is obtained
by solving the equation of the line and the circle simultaneously.
EQUATION OF TANGENTS TO A CIRCLE (at the point (𝒙, 𝒚) on the circle)
Tangents are drawn perpendicular to the radius of the circle from the point the
radius touches the circumference. To obtain the equation of the tangent, we
differentiate the equation of the circle implicitly and thereafter substitute for the
(𝑥, 𝑦) value in the result to arrive at the gradient, 𝑚. After getting the value of the
gradient 𝑚, we then make use of the value of 𝑚 and the (𝑥, 𝑦) value to get the
equation of the tangent to the circle using the 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 − 𝑜𝑛𝑒 𝑝𝑜𝑖𝑛𝑡 𝑓𝑜𝑟𝑚
method of finding the equation of a line.
Examples
1. Find the equation of the tangent to the circle 𝑥 2 + 𝑦 2 − 2𝑥 + 4𝑦 = 15 at the
point (−1, 2)
Solution
Differentiating (implicitly), 2𝑥 + 2𝑦
(2𝑥 + 4)
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
=𝑚=
𝑑𝑦
𝑑𝑥
−2+4
𝑑𝑦
𝑑𝑥
=0
= 2 − 2𝑥
2−2𝑥
2𝑦+4
|(−1,2) =
2−2(−1)
2(2)+ 4
1
= =𝑚
2
Therefore equation of the tangent is
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
1
𝑦 − 2 = (𝑥 + 1)
2
2𝑦 = 𝑥 + 5
2. Find the equation of the tangent to the circle 4𝑥 2 + 4𝑦 2 − 12𝑥 + 24𝑦 = 55
3
at the point (− , 1)
2
Solution
Differentiating (implicitly), 8𝑥 + 8𝑦
(8𝑥 + 24)
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑥
12−8𝑥
8𝑦 + 24
= 12 − 8𝑥
𝑑𝑦
𝑑𝑥
− 12 + 24
𝑑𝑦
𝑑𝑥
=0
𝑑𝑦
𝑑𝑥
3
2
3
12−8(− ,)
2
8(1)+ 24
| (− , 1) =
=
24
32
=
3
4
Therefore equation of the tangent is
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
3
3
4
2
𝑦 − 1 = (𝑥 + )
3
4𝑦 − 4 = 3(𝑥 + )
2
8𝑦 − 8 = 6(𝑥 + 3)
8𝑦 − 6𝑥 = 26
AREA OF A TRIANGLE
The area of a triangle 𝐴𝐵𝐶 whose vertices are 𝐴(𝑥1 , 𝑦1 ), 𝐵(𝑥2 , 𝑦2 ) and 𝐶(𝑥3 , 𝑦3 ) is
given by
Area of ∆𝐴𝐵𝐶 =
1
2
|𝑥1 𝑦2 − 𝑥2 𝑦1 + 𝑥2 𝑦3 − 𝑥3 𝑦2 + 𝑥3 𝑦1 − 𝑥1 𝑦3 |
Examples
1. Find the area of the triangle 𝐴𝐵𝐶, if (3, 4), 𝐵(9, 7) and 𝐶(5, −3)
Using the formula, ∆𝐴𝐵𝐶 =
1
2
|𝑥1 𝑦2 − 𝑥2 𝑦1 + 𝑥2 𝑦3 − 𝑥3 𝑦2 + 𝑥3 𝑦1 − 𝑥1 𝑦3 |
1
We have, |(3.7) − (9.4) + (9. −3) − (5.7) + (5.4) − (3. −3)|
2
th⟹
⟹
1
2
1
2
|21 − 36 − 27 − 35 + 20 + 9| =
. 48 = 24 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠
1
2
|−48|
ADDENDUM TO VECTORS
Component of a vector in the direction of another vector
⃗⃗ is given as 𝐴⃗.
The component of a vector 𝐴⃗ in the direction of vector 𝐵
⃗⃗
𝐵
⃗⃗|
|𝐵
Examples
⃗⃗ = 3𝑖 + 4𝑗. Find the component of
1. Let 𝐴⃗ = 𝑖 + 4𝑗 and 𝐵
i.
⃗⃗
𝐴⃗ in the direction of 𝐵
ii.
⃗⃗ in the direction of 𝐴⃗
𝐵
Solution
i.
⃗⃗| = √32 + 42 = 5
|𝐵
𝐴⃗.
=
ii.
⃗⃗
𝐵
⃗⃗|
|𝐵
= 𝑖 + 4𝑗 .
(𝑖+4𝑗) .(3𝑖+4𝑗)
5
3𝑖+4𝑗
5
⟹
3+16
=
5
19
5
|𝐴⃗| = √12 + 42 = √17
⃗⃗.
𝐵
=
𝐴⃗
|𝐴⃗|
= 3𝑖 + 4𝑗 .
(3𝑖+4𝑗).(𝑖+4𝑗)
5
𝑖+4𝑗
√17
⟹
3+16
=
√17
19
√17
⃗⃗
2. Find the value of 𝑎 such that the component of 𝐴⃗ in the direction of vector 𝐵
⃗⃗ = 𝑖 + 3𝑗.
is equal to 0, if 𝐴⃗ = 𝑎𝑖 + 2𝑗 and 𝐵
Solution
⃗⃗| = √12 + 32 = √10
|𝐵
𝐴⃗.
=
⃗⃗
𝐵
⃗⃗|
|𝐵
= 𝑎𝑖 + 2𝑗 .
(𝑎𝑖+2𝑗) .(𝑖+3𝑗)
√10
Thus, 𝐴⃗.
⃗⃗
𝐵
⃗⃗|
|𝐵
𝑖+3𝑗
√10
⟹
=0⟹
Hence 𝑎 = −6
𝑎+6
√10
𝑎+6
√10
⟹𝑎+6=0
VECTOR PROJECTION
⃗⃗ is given as ⃗Å⃗ =
The scalar projection of a vector 𝐴⃗ unto a vector 𝐵
⃗⃗
𝐴⃗ .𝐵
⃗⃗ |
|𝐵
⃗⃗ is given as
While the vector projection of a vector 𝐴⃗ unto a vector 𝐵
𝑝𝑟𝑜𝑗𝐵⃗⃗ 𝐴⃗ =
⃗⃗
𝐴⃗ .𝐵
⃗⃗ .𝐵
⃗⃗
𝐵
⃗⃗).
(𝐵
Examples
⃗⃗ (ii) ⃗Å⃗
1. Given the vector 𝐴 = 5𝑖 − 𝑗 + 2𝑘 and 𝐵 = 2𝑖 − 𝑗 − 3𝑘. Find (i) 𝑝𝑟𝑜𝑗𝐴⃗ 𝐵
Solution
⃗⃗ =
𝑝𝑟𝑜𝑗𝐴⃗ 𝐵
(i)
⃗⃗ .𝐴⃗
𝐵
𝐴⃗ .𝐴⃗
(𝐴⃗)
⃗⃗ . 𝐴⃗ = (2𝑖 − 𝑗 − 3𝑘). (5𝑖 − 𝑗 + 2𝑘)
𝐵
= 10 + 1 − 6
=5
𝐴⃗ . 𝐴⃗ = (5𝑖 − 𝑗 + 2𝑘). (5𝑖 − 𝑗 + 2𝑘)
= 25 + 1 + 4
= 30
⃗⃗ =
𝑝𝑟𝑜𝑗𝐴⃗ 𝐵
(ii)
⃗Å⃗ =
5
30
5
1
1
6
6
3
(5𝑖 − 𝑗 + 2𝑘) ⟹ 𝑖 + 𝑗 + 𝑘
⃗⃗
𝐴⃗ .𝐵
⃗⃗ |
|𝐵
⃗⃗ | = √22 + (−12 ) + (−3)2 ⟹ |𝐵
⃗⃗ | = √14
|𝐵
⃗⃗ = (5𝑖 − 𝑗 + 2𝑘). (2𝑖 − 𝑗 − 3𝑘) ⟹ 10 + 1 − 6 = 5
𝐴⃗ . 𝐵
⃗Å⃗ =
5
√14