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THE HIGHER MATHEMATICS CLASS

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Presentation on theme: "THE HIGHER MATHEMATICS CLASS"— Presentation transcript:

1 THE HIGHER MATHEMATICS CLASS
WELCOME TO THE HIGHER MATHEMATICS CLASS

2 OBSERVE THE PICTURE CAREFULLY
TELL ME THE SHAPE OF WHEEL.

3 THE CIRCLE CHAPTER -4 EXERCISE -4.1 BOOK BY AKKHOR POTRA

4 LEARNING OUTCOME Equation of circle with center at origin
STUDENT CAN KNOW FROM THIS CHAPTER About the Circle Equation of circle with center at origin Equation of circle with center at the fixed point

5 Necessary formula for circle.
1.Center at origin and radius is r the Equation becomes 2.Center at (h,k) and radius is r the Equation becomes

6 4.Standard Equation of circle is
5.Taking (x1,y1) and (x2,y2) are the extremities of a Diameter of a Circle the Equation of circle is 6.The Equation of circle passes through the points (x1,y1) and (x2,y2) is

7 7.If any circle touches x-axis then radius=ordinate of the center or g2=c
8.If any circle touches y-axis then radius=abscissa of the center or f2=c 9.If any circle touches both axes then center=(radius,radius) 10.The length of chord from x-axis cut by a circle = 11.The length of chord from y-axis cut by a circle =

8 Group Work 1.Find the equation of the circle which passes through origin and intersects 3 &5 unit from the positive direction of x and y axes. X Y

9 Solution: Since the Circle passes through the Origin & intersects 3&5 unit from the positive direction of x & y-axes ,So the Circle will pass through the points (3,0) and (0,5) respectively. Therefore the Equation of Circle becomes (x-3)(x-0)+(y-0)(y-5)+k{(x-3)(0-5)-(y-0)(3-0)}=0 Or,x2-3x+y2-5y+k(-5x+15-3y)=0 … … …(i) Since equ (i) passes through the origin ; so from (i) we get k(0+15-0)=0 Or, 15k=0 Or, k=0 Putting the value of k in equation (i) Therefore, x2+y2-3x-5y=0

10 1.Find the equation of the circle which touches y-axis at(0,3) and passes through the point (-1,0).

11 Solution: Since the Circle passes through the point (0,3) & (-1,0) respectively.
Therefore the Equation of Circle becomes (x-0)(x+1)+(y- 3)(y-0)+k{(x-0)(3 -0)-(y- 3)(0+1)}=0 Or,x2+x+y2- 3 y+k(3 x-y+ 3)=0 Or,x2+y2+x(1+ 3k)-y(3+k)+ 3k=0 … … …(i) Here,

12 Putting the value of k in the Equation (i) we get
x2+y2+x(1+ 3k)-y(3+k)+ 3k=0 … … …(i) Or,x2+y2+x(1+33)-y(3+3)+33=0 Or,x2+y2+x(1+3)-23y+3=0 Or,x2+y2+4x-23y+3=0

13 Evaluation 1.What is Circle? 2.Tell me Characteristics of Circle

14 HOME WORK 1.Find the equation of the circle which passes through the points (3,0),(7,0) and touches y-axis. 2. Find the equation of the circle which passes through the points (3,-1)and touches x-axis at(2,0) . 3.From akkhor potra ex4.1 nos are 1,2,5,7,8,9,10,11.

15 EUCLID ,FATHER OF GEOMETRY
THANKS TO ALL EUCLID ,FATHER OF GEOMETRY


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