SIM
in
Mathematics III
Prepared by:
and
Leonelyn P. Dela Cruz
Rod Umali
Belen A. Bahia
Master Teacher I/Dep’t Coordinator
Domingo N. Viñas
Principal II
Objectives :
At the end of the session , the
selected students will be able to:
1. Identify the difference between a
tangent line and a secant line ;
2. Illustrate the linear and circular
equations by graphing ;
3. Solve systematically and neatly the
point of intersection between a circle
and a tangent line and between a
circle and a secant line.
Let’s discover
the amazing world
of CIRCLES…
8
CIRCLES IN THE COORDINATE PLANE
6
4
2
-10
-5
5
-2
-4
-6
TITLE CARD
-8
10
8
6
4
2
-15
-10
-5
5
-2
-4
-6
-8
10
15
It’s better if we ask our teacher how to
find the point of intersection between
the circle and a tangent line…and
between a circle and a secant line…
let’s go!!!
A line may intersect a
circle at two points, at
one point or at no point
at all…
GUIDE CARD
But before we proceed , you
must know first , how to
identify the equation of the
circle from the equation of the
line.
1.) x + y = 9
equation of the line
2.) x2 + y2 = 100
3.)( x + 4 )2 + ( y – 2 )2 = 52
Equations of
the circle
# 2 is an
equation of a
circle with
center at the
origin and a
radius of 10
You will also use your skills in
Algebra to expand the square
of a binomial…
Ex. (y – 3 )2 = y2 – 6y + 9
To get the
product,
Square the
1st term,
multiply the
product of
the 1st term
and 2nd
term by 2,
then square
the 2nd
term.
…and develop your skills on how to get the
square root of a number ,how to get the product
of a given number raised to a certain exponent,
how to square a binomial ,how to get the factors
of the perfect square trinomial or any given
polynomials …
Oh, I have to
review my
algebra …
…and you must also know how to identify the
radius and the center given the equation of the
circle and vice versa … I know you can do it
by pair…ok?
In the equation
( x – 6 )2 + ( y + 4 )2 = 81,
what is the center?
x–6=0
x=6
y+4=0
y=-4
The center is ( 6 , - 4 )
How about the radius of
that equation?
Easy! Just get the
square root of 81…The
radius is 9…
121…
Yes, I do… the
answers are -11 and
+11…
Do you know
the square
root of 121 ?
How did
you do it?
Just think of a number in
which when you multiply
the number by itself twice,
the product is 121…
54 … just
multiply the base which is 5 ,
four times by itself…4 which
is the exponent , tells us
how many times we multiply
the base by itself… so,
54 = 5x5x5x5
= 25x25
= 625…
She’s studious
and diligent in
the class, I will
ask her…
Hi, Ms. Butterfly,
I’m Spider, your
classmate in
Geometry… Can
you help me find
the value of 54 ?
Thank
you…
You’re
welcome…
Penguin,
do you
know the
factors of
t2 +2t - 35?
Get the square root of the 1st
term, then think of the
factors of the 3rd term, such
that when you add them, the
result is the numerical
coefficient of the 2nd term…
The answer is
( t +7 ) ( t - 5 )
Are you sure
with your
answer?
Yes, I’m sure
with my
answer…
Can you help
me find the
product of
( 3x – 4 )2?
Ok, just square the 1st term, multiply the product
of the 1st & 2nd term by 2, then square the second
term… or use the happy face method…
( 3x – 4 )2 = ( 3x – 4 ) ( 3x – 4 )
= 9x2 – 12x – 12x + 16
( 3x – 4 )2 = 9x2 – 24x + 16
Ok, thanks…
I’ll practice it at
home...
You’re
welcome…
Earlier, you ask me how to solve for the point of
intersection of the circle and a tangent line. Here, I
will give you two equations like
eq.1
x2 + y2 = 20
eq. 2
x – 2y = 10
We have to use eq. 2 to solve for x…
x2 + y 2= 20
x – 2y = 10
eq.1
eq. 2
x – 2y= 10
x = 2y+ 10
So, the value of
x is 2y + 10
…to solve for y,
substitute the value
of X in eq. 1….
X = 2y + 10
x2 + y2 = 20
(2y + 10)2 + y2 = 20
4y 2+ 40y + 100 + y2 = 20
5y2 + 40y + 80 = 0
5 ( y2 + 8y + 16 ) = 0
5 (y + 4 )2 = 0
y = -4
…so, the value of y = -4
Simplify(2y + 10 )2, it
becomes 4y2 + 40y + 100.
…then
add
similar
terms…
…take
out the
common
factor…
…get the factors of
(y2 + 8y + 16 )… we have
( y + 4 )2…
To solve for x, substitute
the value of y in eq. 2
y = -4
x – 2y = 10
x– 2( -4 ) = 10
x + 8 = 10
x = 10-8
x= 2
ss: ( 2 , -4)
(x,y)
…the solution set is ( 2 ,- 4 ).
…here is the graph of the equations:
x2 + y2 8= 20 and x – 2y = 10…The circle
and the line intersect at point ( 2 , - 4)…
...Do you have any clarifications?... If
6
none, let’s proceed to example # 2…
44
Y
x2 + y2 = 20
3
22
1
-5
-5
-4 -3 -2 -1
0
-1
1
2
3
4
55
6
X
10
-2-2
-3
-4-4
-5
-6
x– 2y = 10
A
( 2, - 4 )
15
Our next example will teach you how to solve
for the point of intersection of the circle and a
secant line, here are the equations:
eq.1
x2 + y2 = 25
eq. 2
x–y+1=0
We will solve for x using eq.2 …
x2 + y 2= 25
x–y +1=0
eq.1
eq. 2
x–y+1=0
x=y-1
So, the value of
x is y - 1
…substitute the
value of X in eq. 1…
x = y -1
+
= 25
(y - 1)2 + y2 = 25
y 2- 2y + 1 + y2 = 25
2y2 - 2y -24= 0
2( y2 - y - 12) = 0
2 ( y + 3 ) (y - 4 ) = 0
y = -3
x2
y2
y=4
…simplify ( y - 1 )2, it
becomes y2 - 2y +1…
…then
add
similar
terms…
…take
out the
common
factor…
…get the factors of
(y2 - y - 12)…these are:
( y + 3 ) and ( y – 4 )
…there are two values
of y… we have , y = -3
and
y = 4…substitute
them to eq. 2…
If y = -3
x= y–1
x= -3 – 1
x = -4
ss: ( -4, -3)
(x,y)
…It means that the intersection
of the circle x2 + y2 = 25 and a
line x – y + 1 = 0 is at (-4,-3) and
(3,4)… That means, the line is a
secant line because it touches
the circle at two points…
If y = 4
x=y-1
x= 4-1
x=3
ss: ( 3 , 4)
(x,y)
…any clarifications with Ex. # 2 ?...
8
…If none, here’s
the graph of x2 + y2 = 25 and
x –y + 1 = 0…The circle and the line intersect at
point ( - 4 ,- 3 ) and ( 3,4 )…
6
x2 + y2 = 25
Y
5
4
( 3,4 )
4
3
2
2
1
-5 -4 -3 -2
-5
-1 0
-1
-2-2
(-4 , -3)
-3
-4-4
-5
x– y + 1 = 0
-6
1
2
3
4
55
X
10
ACTIVITY CARD
Activity 1
When the clock strikes 13, what time is it?
To find the answer to this riddle , identify the center and the radius of
each equation, then use the decoder to reveal the answer.
1.) x2 + y2 = 121
GET
2.) ( x – 4 )2 + y2 = 16
IT
3.) x2 + ( y – 3 )2 = 10
TIME
4.) x2 + y2 + 4x – 6y = - 4
TO
5.) x2 + y2 – 10x – 6y = - 18
FIXED
C ( 0 , 3 ) r= 10
C ( -2, 3 ) , r = 3
C ( 0 , 0 ) , r =11
C(4,0),r=4
C(5,3)r=4
DECODER
1
2
3
4
5
Activity 2
What did the circle say to the tangent line?
You can answer this by helping the four mice to the right path in a maze
and collect all the letters that corresponds to the equations inside the
maze and decode it to the boxes below.
1. C ( 1 , 2 ) , r = 4
5. C ( -5 , 2 ) , r = 10
2.C(0,0) , r=8
6. C ( 4 , -2 ) , r = 7
3. C ( 4 , -7 ) , r = 3
7. C ( 0 , 0 ) , r = 2
4. C ( -2 , -5 ) , r = 11
2
MAZE
( x - 1 )2 + ( y – 2 )2 = 16
x 2 + y2 = 64
x 2 + y2 - 8x + 14y = -56
x 2 + 4x + y 2 + 10y = 92
x2 + 10x + y 2 - 4y = 71
x 2 + y 2- 8x + 4y = 29
x2+y2=8
x 2 + 4x + y 2 + 10y = 92
UC
NG
OP
ME
ST
HI
TO
x 2 + y 2- 8x + 4y = 29
x 2 + y2 = 64
( x - 1 )2 + ( y – 2 )2 = 16
x2 + 10x + y 2 - 4y = 71
x 2 + y2 - 8x + 14y = -56
x2+y2=8
1
2
3
4
5
6
7
Find the intersection of a line and a circle.
1.) x2 + y2 = 9
x+y=3
4. ) x2 + y2 = 100
X=6
2.) x2 + y2 = 16
x-y=4
3.) x2 + y2 = 25
y=3
ASSESSMENT CARD # 1
(1 )
x2 + y2 = 29
y–x=3
(2)
x2 + y2 = 169
y– x = 7
(5)
x2 + y2 = 225
x + y = 15
ASSESSMENT CARD # 2
(4)
(3)
x2 + y2 = 400
x + y = 20
x2 + y2 = 25
2y = 10
Student’s Note
Please check the circles below:
After reading this lesson, I now understand the
following concepts:
Tangent line intersects the circle at exactly one point;
Secant line intersects the circle at two distinct points;
How to find the point of intersection between a circle
and a tangent line;
How to find the point of intersection between a circle
and a secant line;
Finding the radius and the center given the equation
of the circle in which the center is found at the origin;
Finding the radius and the center given the equation of
the circle in which the center is not found at the origin;
Finding the equation of the circle given the radius and
the center;
Writing the general form of the equation of the circle
in center-radius form.
Student’s Note
The activities have been:
Easy
Average
Difficult
Now, I can do the activities :
Alone
With one of my
classmates to work with
Within a group
ENRICHMENT CARD
Theater Lighting
A bank of lights is arranged over
a stage. Each light illuminates a
circular area on the stage. A
coordinate plane is used to arrange
the lights, using the corner of the
stage as the origin . The equation
( x – 13 )2 + ( y – 4 )2 = 16 represents
one of the disks of light.
a. Graph the disk of light.
b. Three actors are located as
follows : Leonel is at ( 11 , 4 ),
Rod is at ( 8 , 5 ) and Philip is at
( 15 , 5 ) . Which actors are in the
disk of light?
CELL PHONES
A cellular phone network uses towers
to transmit calls. Each towers transmits to
a circular area. On a grid of a city, the
coordinates of the location and the radius
each tower covers are as follows
( integers represent kilometers): Tower A is
at ( 0 , 0 ) and covers a 3 km radius, Tower
B is at ( 5 , 3 ) and covers a 2.5 km radius,
and Tower C is at ( 2 , 5 ) and covers a 2 km
radius.
a. Write the equation s that represent the
transmission boundaries of the two
towers . Graph each equation.
b. Tell which towers , if any, transmit to a
phone located at J ( 1 , 1 ) , K ( 4 , 2 ),
L ( 3.5 , 4.5 ) M ( 2 , 2.8 ) , or N ( 1 , 6 ).
THE ANSWER KEY CARD
Answer for Activity Card # 1
TO
TIME
IT
GET
FIXED
Answer for Activity Card # 2
S
T
1
O
P
2
T
O
3
U
C
4
H
I
5
N
G
6
M
E
7
Assessment Card # 1
1
2
3
4
Assessment Card # 2
1
( -5 , -2 ) and ( 2 , 5 )
(-12,-5 ) and ( 5 , 12 )
( 4 , 0 ) and ( 0 , -4 )
2
( 0, 5 )
( 4 , 3 ) and (-4 , 3 )
3
4
( 0 , 20 ) and ( 20 , 0 )
( 3 , 0 ) and ( 0 , 3 )
( 6 , 8 ) and ( 6 , -8 )
5
( 0 , 15 ) and ( 15 , 0 )
Answers for Enrichment Card # 1
a. Rewrite the equation to find the center and radius:
( x – 13 )2 + ( y – 4 )2 = 16
( x – 13 )2 + ( y – 4 )2 = 42
The center is ( 13, 4 ) and the radius is 4. The circle is shown below.
Y
8
66
Philip
( 15 , 5 )
Rod
(8,5)
44
Leonel
( 11 , 4 )
( 13 , 4 )
22
5
5
-2
10
10
15
15
X
20
b. The graph shows that Leonel and Philip are both in the disk of
light.
Answers for Enrichment Card # 2
a. Tower A ( x2 + y)2 = 9
Tower B ( x – 5 )2 + ( y – 3 )2 = 6.25
Tower C ( x – 2 )2 + ( y – 5 )2 = 4
Y
8
6
b. J ( 1, 1 ) – Tower A
K ( 4 , 2 ) – Tower B
L ( 3.5 , 4.5 ) – Tower B & C
M ( 2 , 2.8 ) – No signal
N ( 1 , 6 ) – Tower C
L
4
4
B
3
M
-5
K
2
J
1
X
C
5
2
-10
N
6
A
0
-2
-4
1
2
3
4
5
5
BOOKS
R
E
F
E
R
E
N
C
E
C
A
R
D
Geometry, Revised Edition (NPSBE)
Dilao , Soledad J. et al, SD Publications , Inc.
Quezon City, 2009
Geometry ( Applying , Reasoning, Measuring )
Larson, Ron et al, McDougal Littell , Illinois, 2004
Work Text in Geometry Simplified Concepts and Structures
Pascual, Ferdinand C. et al , Innovative Educational
Materials , Inc, Manila, 2002
Geometry, Teacher’s Edition
Clemens , Stanley R. et al, Addison-Wesley Publishing
Company, Inc. , USA, 1994
SOFTWARE
Dynamic Geometry Software for Exploring Mathematics
Version 4.06, KCP Technology, Inc. 2001
WEBSITE
http://www.classzone.com
http://www.mcdougalittell.com
http://www.keypress.com/sketchpad