Download Theorem of tangent

Useful formula :
p
 log 10 p  log 10 q
q
log 10 pq  log 10 p  log 10 q
log 10
log 10 p n  n log 10 p
a  b 2  a 2  2ab  b 2
(a  b)(c  d )  a(c  d )  b(c  d )
( x  m)( x  n)  x 2  mx  nx  mn
 ac  ad  bc  bd
 x 2  (m  n) x  mn
a  b 2  a 2  2ab  b 2
a 2  b 2  a  b a  b 
a 2  2ab  b 2  a  b 
a 2  2ab  b 2  a  b 
2

2
a 3  b 3  a  b a 2  ab  b 2


a 3  b 3  a  b a 2  ab  b 2

ax 2  bx  c  0
General form of quadratic equation:
Quadratic formula
 b  b 2  4ac
x
2a
Nature of roots:
ax 2  bx  c  0
discriminant:
Item
Description
Nature of
roots
0
0
0
  b 2  4ac
Two unequal real
roots
One real root
No real root
E.g.
Find the value for k if the equation x  kx  9  0 has two
distinct real roots
Solution
  b 2  4ac  0
( k ) 2  4(1)( 9)  0
2
Explanation
y  k 2  36
k  36  0
2
(k  6)( k  6)  0
-6
6
k < -6 or k > 6#
E.g.
Solve x  x  0 .
Solution
x  x2  0
Explanation
2
x x0
2
x( x  1)  0
0 < x < 1.
y  x2  x
Find the roots of x  x  0 by
factorization.
2
The region of the curve that is below zero
is between 0 and 1.
Ref.:  at centre twice  at ⊙ce
Ref.: line from centre  chord bisects chord
Ref.: line joining centre to mid-pt. of chord  chord
Ref.: equal chords, equidistant from center
Ref.: chords equidistant from centre are equal
Ref.: tangent  radius
Theorem of tangent
If two tangents are drawn to a circle from an external point,
(a) the tangents are equal;
(b) the tangents subtend equal angles at the centre;
(c) the line joining the external point to the centre bisects the angle between the tangents.
i.e.
If TA, TB are tangents from T,
A
then
TA = TB;
and
TOA = TOB; and
ATO = BTO
O
Ref.: tangent properties
T
B
Remainder Theorem states that:
When a polynomial f(x) is divided by a linear divisor x – a, the remainder R is equal to f(a).
i.e.
Remainder
R = f(a).
Factor Theorem states that:
x – a is a factor of the polynomial f(x) if and only if f(a) = 0.
When the ratio of the two quantities x and y is a:b
i.e. x:y=a:b,
x a

y b
x y
  k for some constant k
a b
or x=ka and y=kb
Similarly, if three quantities x, y and z of the same kind are to compare and that
x y z
x:y:z=a:b:c, then  
a b c
or x=ka, y=kb and z=kb for some constant k.
If a : b = c : d, then a, b, c and d are said to be in proportion.
If two quantities x and y are so varying that the ratio of y to x is always a constant, then y
is said to vary directly as x. The relation may be written y  x and read as ‘y is
proportional to x’ or ‘y varies directly as x’.
Hence if y  x
y
then
= k (k = constant)
x
or y = kx.
A quantity y is said to vary inversely as or is inversely proportional to another quantity
1
x if y varies directly as .
x
In symbol, if y 
k
1
then y  where k is the variation constant,
x
x
or xy = k.
Hence when y varies inversely as x, the product xy is a constant
A quantity is said to vary jointly as several other quantities, if it varies as the product of
these quantities.
z varies jointly as x and y, if z = k xy where k is the variation constant.
A joint variation can also be a combination of direct and inverse variations. If a varies
b
directly as b and inversely as c, then a=k where k is the variation constant.
c
A quantity z may be composed of several parts. One of the parts varies as another
quantity x and a second part varies as a third quantity y and so on. We call this kind of
variation a partial variation and the equation involves a sum of several parts.
For example, if z varies partly directly as x and partly directly as y, then
z = k1x + k2y
where k1 and k2 are variation constants.
If z is partly constant and partly varies directly as x, then
z = k1 + k2x
where k1 and k2 are variation constants.
=  : 2
L : 2r
L = r
In figure 2, denote A = area of a sector
From proportionality, we have
area of sector:
area of circle = angle at centre: angle of a full circle
i.e.
A : r2 =  : 2
A
1 2
r
2
Figure 2

Also, by considering L = r , A can be expressed as
A
1
rL
2
Area of a triangle =
the Sine formula :
1
ab sin C
2
a
b
c


sin A sin B sin C
the Cosine formula :
c 2  a 2  b 2  2ab cos C
or
cos C 
a2  b2  c2
2ab
An arithmetic progression is a progression in which any term minus its previous terms is a
constant. i.e. T(2) - T(1) = T(3) – T(2) = T(4) – T(3) = … = constant. The definition can be
written as
T (n  1)  T (n )  constant
(F1)
 The constant is called the common difference, d.
d  T (n  1)  T (n)
(F2)
 a is used to represent the first term.
The general term of any AP can be written as
T ( n )  a  ( n  1)d
Arithmetic Series
A series is a sum of term. 2+4+6+8+10+12+14 is a series, 3+7+5+9 is a series.
We use the S(n) to represent the sum of n term: S(n) = T(1) + T(2) + T(3) + … + T(n).
For an arithmetic progression,
n
S (n)  2a  (n  1)d 
(F5)
2
if we use  to represent the last term, T(n),
n
S ( n)  ( a   )
(F6)
2
both of the formula are useful.
A geometric progression is a progression in which the ratio of each term to the preceding term
T (2) T (3) T (4)
is a constant. i.e.


 ...  constant . It can be written as
T (1) T (2) T (3)
T (n  1)
(F1)
 constant
T ( n)


The constant is called the common ratio, R.
T (n  1)
R
T ( n)
a is used to represent the first term.
The general term of any geometric progression can be written as
T (n)  aR n1
3.
(F2)
(F3)
Geometric Series
For a geometric progression, the series T (1)  T (2)  T (3)    T (n ) is given by
S ( n) 
a(1  R n ) a( R n  1)

1 R
R 1
The sum to infinity of the geometric series S() is given by
a
S ( ) 
( –1 < R < 1)
1 R
(F5)
(F6)
Probability of an event E
For a random experiment in which every outcome is equally possible, the probability
of an event E is defined as :
P(E ) 
Number of outcomes in the event E
number of facourable outcomes

Number of outcomes in S
Total number of outcomes
so the probability for either E or F to occur is :
P(E or F) = P(E  F) = P ( E ) + P ( F )
For two independent events E and F,
P ( E and F ) = P ( E ) x P ( F )
P(A&B) = P(A)xP(B after A has occurred)
Grouped frequency distribution
The steps of constructing a grouped frequency distribution are as follows:
Step 1:Construct the classes
a.
Pick out the highest value and the lowest value and find the range of the data.
b.
Determine the class intervals. Number of intervals should be between 5 and 12 and they
usually have equal widths.
c.
Make sure that each item of the data set goes into one and only one class.
Step 2:Tally the data into these classes.
Step 3:Total the tallies in each class to give the class frequency.
Mean( x) 
x1 f1  x2 f 2  x3 f 3  ...  xn f n
f1  f 2  f 3  ...  f n
Median
For ungrouped data,
Median = the middle datum, when n is odd.
Median = the mean of the two middle data, when n is even.
For ungrouped data, mode is the datum that has the highest frequency.
For grouped data, modal class is the class that has the highest frequency.
Inter quartile range
Inter quartile range = Q3 – Q1
where Q1, Q2, Q3 are called quartiles which divide the data (which have been ranked, i.e. arranged in order) into
four equal parts.
Moreover,
Q2 is the median of the whole set of data,
Q1 is the median of the lower half,
Q3 is the median of the upper half.
Quartile deviation, Q.D. = ½ (Q3  Q1)
Standard deviation
For ungrouped data x1, x2,…,xn, with a mean
x , the standard deviation () is
( x1  x ) 2  ( x 2  x ) 2  ( x 3  x ) 2  ...  ( x n  x ) 2
n
For grouped data with class marks x1, x2,…,xn; corresponding frequencies f1,f2,…,fn, and a mean
deviation () is
( x1  x ) 2 f 1  ( x 2  x ) 2 f 2  ( x 3  x ) 2 f 3  ...  ( x n  x ) 2 f n
f 1  f 2  f 3  ...  f n
x , the standard
For a distribution of marks with mean
The standard score z is
z
x and standard deviation ,
xx

A normal curve has the following characteristics:
1.
It is symmetrical about the mean.
2.
Mean = mode = median. They all lie at the centre of the curve.
3.
There are fewer data for values further away from the mean.
a) about 68% of the data lie within 1 standard deviation from the mean.
b) about 95% of the data lie within 2 standard deviations from the mean.
c) about 99.7% of the data lie within 3 standard deviations from the mean.
Section formula
A point P on the joining of AB is said to divide AB in the ratio r 1 : r2 if AP : PB = r1 : r2. In the figure below,
if P divides AB in the ratio r1 : r2
Hence, the section formula concludes that:
The coordinates of the point P which divides the joining of A(x1,y1) and B(x2,y2) is
x
r1 y 2  r2 y1
r1 x 2  r2 x 1
and y 
r1  r2
r1  r2
The slope of a line is defined as the tangent of the inclination of the line.
i.e.
m = tan 
Therefore, the equation of the line joining (x1,y1) and (x2,y2) is
y  y1 y 2  y1

x  x1
x2  x1
Point-slope form
We have already found in the last section that the line through the point (x1,y1) of slope m is:
y  y1
m
x  x1
The equation of the line with slope m and y-intercept b is the line through (0,b), so its
equation is:
y b
 m or
x0
y=mx+b
The stand form of a circle is ( x  h) 2  ( y  k ) 2  r 2 , expanding this equation gives
… … (*)
x 2  y 2  2hx  2ky  (h 2  k 2  r 2 )  0
From this and the results we have worked out in the previous examples, we realize that the
equation of a circle is quadratic in both x and y but without xy term. Besides, the
coefficients of x2 and y2 terms are equal (usually set to unity). We therefore conclude that
every circle can be written in the form
x 2  y 2  dx  ey  f  0 … … (**)
and (**) is called the general form of a circle.
If (*) and (**) represent the same circle, then
d  2h

e  2k
 f  h2  k 2  r 2


h 


or k 


r 

d
2
e
2
d
e
( )2  ( )2  f
2
2
Hence x 2  y 2  dx  ey  f  0 represents a circle centered at (
d
e
r  ( )2  ( )2  f .
2
2
Good Luck to You !
- The End -
d e
,
) with radius
2 2