Analytical Geometry Circle Parabola Ellipse Hyperbola The Circle - - PowerPoint PPT Presentation

Analytical Geometry

• Circle
• Parabola
• Ellipse
• Hyperbola

The Circle

Definition

The circle is the locus of a point moving such that its distance from a fixed point (the center) is constant (the radius).

Equation of a circle

The equation of a circle with center at and radius r is:

( , ) x y

2 2 2

( ) ( ) x x y y r    

( , ) c x y r

Example:

(i) with center (2, 3) and radius 5.

Solution

(iii) with (2,3) and (4,-5) represent two end points of a diameter. Find the equation of circle (ii) with center (-4, 3) and passes through (-1, -1). (iv) with center (-1, -4) and tangent to x-axis. (i) The equation is

   

2 2 2

2 3 5 x y    

2 2

4 6 12 x y x y     

(ii) with center (-4, 3) and passes through (-1, -1). Radius equal the distance between the center and any point on the

• circle. So

   

2 2 1 1

r x x y y    

   

2 2

4 1 3 1 5 r      

   

2 2 2

4 3 5 x y    

(iii) with (2,3) and (4,-5) represent two end points of a diameter. The center is the mid point of ends of diameter

   

2 2

2 3 3 1 17 r     

1 2 1 2

, 2 2 x x y y Center         

   

2 2

3 1 17 x y    

 

2 4 3 5 , 3, 1 2 2 Center           

(iv) with center (-1, -4) and tangent to x-axis. 4 So r =4

   

2 2

1 4 16 x y    

The General Equation of a Circle

The general equation of a circle can be written in the form:

2 2

2 2 x y fx gy e     

with center at (-f, -g) and radius

2 2

r f g e   

(i) Coeff. of = coeff. of . (ii) If r > 0 then, we have a real circle (iii) If r < 0 then, we have an imaginary circle. (iv) If r = 0 then, we have a point circle (circle with radius zero).

2

x

2

y

Example:

Does the equation represent a real circle? If so, find the center and the radius of this circle.

7 y 8 x 4 y 2 x 2

2 2

    

Solution

The equation of the circle is reduced to

2 2

2 4 7/ 2 x y x y       f=1, g =2 and e = -7/2

1 4 7/2 2.915 r     

we have a real circle The center (-f, -g) = (-1,-2)

Example:

Find the equation of a point circle with center at (2, -1).

Solution

Point circle

r  

Then the equation of the circle is

2 2

( 2) ( 1) x y     

2 2

4 2 5 x y x y     

Example:

Find the equation of a circle that passes through the three points (1, 2), (0,3) and (0,-3).

Solution

General equation of circle is:

2 2

2 2 x y f x g y e     

Substitute with the three points (1, 2), (0,3) and (0,-3) into the equation of the circle, we get the following three equations:

2 4 5 f g e     6 9 g e    6 9 g e    

(1) (2) (3)

After solving equations (2) & (3) we get:

9 e   g 

Then substitute in equation (1) we get:

2 f 

And the equation of the circle will be:

2 2

4 9 x y x    

Example:

Find the equation of a circle that passes through the two points (-1, 2), (-4,3) and the center lies on the line 4x-3y = 5.

Solution

General equation of circle is:

2 2

2 2 x y f x g y e     

Substitute with the three points (-1, 2), (-4,3) into the equation

• f the circle, we get the following two equations:

2 4 5 f g e      8 6 25 f g e     

(1) (2)

After solving equations (1), (2) & (3) we get:

7 f  11 g 

35 e  

And the equation of the circle will be:

2 2

14 22 35 x y x y     

4 3 5 f g   

(3)

Substitute with the center coordinates (-f,-g) into the equation

• f the line 4x-3y = 5, we get: