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MATHEMATICS TASK

BY :

XI – IPA – 6 SMAN 1 PURWOKERTO

SEKOLAH MENENGAH ATAS NEGERI 1 PURWOKERTO

TAHUN AJARAN 2009-2010



MATHEMATICS TASK

CIRCLE

BY :

1. ANGGRAENI DYAN PITALOKA XI-IPA-6 / 05

2. DWI ANITA NUR FITRIANI XI-IPA-6 / 11

3. FELIA AGUSTINA XI-IPA-6 / 14

4. GITA RATRI PRAFITRI XI-IPA-6 / 15

5. MAWARWATI DYAH XI-IPA-6 / 25

6. PUPUT ANIS SOLIKHAH XI-IPA-6 / 29

7. SUFRAHA ISLAMIA XI-IPA-6 / 38

SEKOLAH MENENGAH ATAS NEGERI 1 PURWOKERTO

TAHUN AJARAN 2009-2010



CIRCLE

• T HE CIRCLE WITH CENTER OF O (0,0)

The locus of points that is equal to a certain point. The same distance

to the circle is called the radius of the circle, whereas certain point called thecenter of the circle.

Based on the definition of the circle above, it will obtain the equation

of the circle of radius r and center have O (0.0):

The point T (x,y) is the circle with center of O(0,0) and radius r. By

definition, the position of point T toward the circle:

{T(x,y) TO= r}

{T(x,y) TO2= r 2 }

{T(x,y) (x-0) 2 + (y-0) 2 = r 2 }

{T(x,y) x2 + y2 = r 2 }

So, the equation of circle with center of O(0,0) and radius of r is

x2 + y2 = r2

FORMULA:

x2 + y2 = r2

Example:



1. Determine the equation of the circle with center of O(0,0) and radius of

3.

Answer :

x2 + y2 = r 2

x2 + y2 = 32

x2 + y2 = 9

Equation of the circle of radius and 4 cm centered at (0.0) are:

x2+ y2 = 9.

2. Determine the equation of the circle with center of O(0,0) and radius of

4.

Answer :

x2 + y2 = r 2

x2 + y2 = 42

x2 + y2 = 16

Equation of the circle of radius and 4 cm centered at (0.0) are:

x2+ y2 = 16.

• GENERAL FORM OF CIRCLE EQUATION

x2

+ y2

+ Ax + By + C = 0

Center : B A2

1,

2

1

−−



Radius :

Example :

1.Determine the center from this equation

x2+ y

2+ 6x -10y + 8=0

Answer :

center : [-1/2(6),-1/2 (-10)]

= (-3,5)

2. Determine the center and radius from this equation x2+ y2 -4x +6y -12=0

Answer :

center : [-1/2 (-4),-1/2(6)]

= (2,-3)

radius :

=

COMPETENCE CHECK

1. Given a point: P (-2.0); Q (0.1), R (5.2) and S (0.8). Point which lies

outside the circle with the equation below is ...

A. PB. Q

C. P and Q

D. R and S

E. P, Q and S

C B A −+ 2^4

12^

4

1

1294 −+

1



2. The coordinates of the center and radius of the circle with the equation

below is ... ,

A. Center (-3, -1) and the radius 2

B. Center (-1, -3) and radius 2

C. Center (3.1) and radius 2

D. Center (3.1) and the radius of 4

E. Center (1.3) and the radius of 4

3. Equation of the circle of radius and 8 cm centered at (0.0) are:

4. Equation of the circle of radius and 3,5 cm centered at (0.0) are:

5. Determine the center and radius from this equation x2+ y2- 4x -12=0 is…

6. Determine the center and radius from this equation x2+ y2 +6y=0 is…

• THE CIRCLE WITH CENTER OF P(A,B)



The circle’s equation with center of P(a,b)

(x-a)2 + (y-b)2 = r2

Example:

1. The circle’s equation with center of P(2,3) through point of A(-2,0) is…

Answer :

r = PA

=

=

=

=

= 5

So the circle equation is

= (x-2)2 + (y-3)2 = 52

= (x-2)2 + (y-3)2 = 25

2. The circle’s equation with center of P(-5,3) and radius r = is…

Answer :

r =

= 4 x 3

= 12

So the circle equation is



(x+5)2 + (y-3)2 = 12

• THE DISTANCE BETWEEN CENTER P ( X1 , Y1) TO THE

LINE OF AX + BY + C = 0

The Distance Between Center P ( X1 , Y1) to the line of ax + by + c = 0 is

r =

• Example :

1. Determine the circle’s equation with center of P ( 2 , 3 ) and has tangent line

of 3x + 4y + 2 = 0.

Answer :

Given the circle with the center P ( 2 , 3 ) and has line of 3x + 4y + 2 = 0, as

its tangent line. Thus, radius (r) of the circle is the distance of point P

( 2 , 3 ) to the line of 3x + 4y + 2 = 0.

From point of P (X1,Y1) to the line of ax + by + c = 0 is formulated as

r =

then r =

=

=



= 4

So the circle’s equation is ( x – 2 )2 + ( y – 3 )2 = 16

2. Determine the circle’s equation with center of P (-1,0) and has line of 3x +

4y - 12 = 0 as its tangent line.

Answer :

Given the circle with the center P (-1,0) and has line of 3x + 4y - 12 = 0, as

its tangent line. Thus, radius (r) of the circle is the distance of point P (-1,0)

to the line of 3x + 4y - 12 = 0.

From point of P (X1,Y1) to the line of ax + by + c = 0 is formulated as

r =

then r =

r =

r =

r =

r =

r= 3



So the circle’s equation is ( x + 1 )2 + y 2 = 9

• THE EQUATION OF THE CIRCLE WITH

DIAMETER OF AB

Formula :

Since AB is diameter, then the center of the circle is at the midpoint of AB. To

calculate the coordinate of the center, we can use the formula of

P

Example :

1. Determine the equation of the circle with diameter of AB where A(-3,2) and

B(3,-4)!

Answer :

P (0,0)

Since the center is P (0,0) and through point of A, as well as point of B, then the

formula of a circle is X2 + Y2 = r2. To calculate the value of r we can utilize both

of point A or B

X2 + Y2 = r 2 (if suppose we use coordinate of point A (-3,4)

(-32) + 42 = r 2

r 2 = 25

Thus, the circle’s equation is X2 + Y2 = 25



2. Determine the circle’s equation which has diameter of AB where A(3,-2) and

B(1,6)!

Answer :

P (0,0)

Since the center is P (2,-2) and through point of A, as well as point of B, then the

formula of a circle is X2 + Y2 = r 2.

X2 + Y2 = r 2

(22) + (-2)2 = r 2

r 2 = 8

Thus, the circle’s equation is

X2 + Y2 = 8

• TANGENCY WITH X-AXIS AND TANGENCY WITH

Y-AXIS

Formula :

Tangency with X-axis :

(x-a)2 + (y-b)2 = b2

Tangency with Y-axis :

(x-a) 2 + (y-b)2 = a2

Example :



1. The circle in P (-2,4) with tangency in x-axis

Answer :

(x-a)2

+ (y-b)2

= b2

(x+2)2 + (y-4)2 = 42

r = 6

2. The circle in P (1,-3) with tangency in y-axis

Answer :

(x-a)2 + (y-b)2 = a2

(x-1)2 + (y+3)2 = 12

r = 1

• THE POSITION OF POINT IN CIRCLE

Formula :

1.Circle x²+y²=r²

a) If point of (x1,y1) at the circle x2 + y2 =r 2,so that x12 + y1

2 =r 2

b) If point (x1,y1) inside the circle x2 + y2=r2,so that x12+y1

2<r 2

c) If point (x1,y1) outside the circle x2+y2=r 2,so that x12+y1

2>r 2

2. Circle(x-a)2 + (y-b)2 =r 2

a)If point (x1,y1) at the circle (x-a)2+(y-b)2=r 2,so that

(x1-a)2+(y1-b)2=r 2

b)If point (x1,y1) inside the circle (x-a)2+(y-b)2=r 2,so that

(x1-a)2+(y1-b)2<r 2

c)If point (x1,y1) outside the circle (x-a)2+(y-b)2=r 2,so that

(x1-a)2+(y1-b)2>r 2

Example :

1. Determine position of point P(2,3), Q(4,-2), and R(5,8)



Answer:

x2 + y2 -8x-2y+8=0

(x-4)2

+ (y-1)2

=32

P(2,3) → (2-4)2+(3-1)2=8<9

Q(4,-2) → (4-4)2+(-2-1)2=9

R(5,8) → (5-4)2+(8-1)2=50>9

So,point of P is inside the circle,point of Q is at the circle,and point of R

is outside the circle.

2. Determine the point of P (2,3), Q (4,-2), and R (5,8) with equation

x2 + y2 – 8x – 2y + 8 = 0

Answer :

x2 + y2 – 8x – 2y + 8 = 0

(x – 4) 2 + (y – 1) 2 = (3) 2

P (2,3) → (2 – 4) 2 + (3 – 1) 2 = 8 < 9

Q (4,-2) → (4 – 4) 2+ (-2 - 2) 2 = 9

R (5,8) → (5 – 4)2

+ (8 – 1)2

= 50 > 9

So, the point P is inside the circle, the point Q is at the circle, and the

point R is outside the circle



COMPETENCE CHECK

1..Determine the circle’s equation which has center on O(0,0) with has a

diameter of AB where A(-3,2) and B(3,-2)!

2. The equation of the circle with center at ( -2,4 ) and has line of 3x – 4y – 3

= 0 is……………

3. Determine the circle’s equation with center of P(3,3) and radius 3 is…

4. Determine the position of point T(x,y) so that{ T(x,y) | TA =3TB} if A(9,0)

and B(1,0) ……………

5. Determine the point of P(3,4) and Q(6,-1) with equation x2 + y2 – 8x – 2y +

8 = 0……………

6. The circle in P (-6,8) with tangency in x-axis……………

7. The circle in P (4,4) with tangency in y-axis……………

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