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 r moving such that its distance from a c x ( , y ) fixed point (the center) is constant 0 0 (the radius). Equation of a circle ( x , y ) The equation of a circle with center at and 0 0 radius r is:     2 2 2 ( ) ( ) x x y y r 0 0

Example: Find the equation of circle (i) with center (2, 3) and radius 5. (ii) with center (-4, 3) and passes through (-1, -1). (iii) with (2,3) and (4,-5) represent two end points of a diameter. (iv) with center (-1, -4) and tangent to x-axis. Solution      2   2  2 x 2 y 3 5 (i) The equation is      2 2 x y 4 x 6 y 12 0

(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 r x x y y 0 1 0 1     r       2 2 4 1 3 1 5      2   2  2 x 4 y 3 5

(iii) with (2,3) and (4,-5) represent two end points of a diameter. The center is the mid point of ends of diameter     x x y y   1 2 1 2  Center ,   2 2     2 4 3 5        Center , 3, 1   2 2     r      2 2 2 3 3 1 17      2   2  x 3 y 1 17

(iv) with center (-1, -4) and tangent to x-axis. So r = 4 4      2   2  x 1 y 4 16

The General Equation of a Circle The general equation of a circle can be written in the form:      2 2 x y 2 fx 2 gy e 0    2 2 r f g e with center at (-f, -g) and radius 2 x 2 y (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).

Example: Does the equation represent a real circle?      2 2 2 x 2 y 4 x 8 y 7 0 If so, find the center and the radius of this circle. Solution The equation of the circle is reduced to       f=1, g =2 and e = -7/2 2 2 x y 2 x 4 y 7/ 2 0 r      1 4 7/2 2.915 0 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   0 r Point circle Then the equation of the circle is      2 2 ( 2) ( 1) 0 x y      2 2 x y 4 x 2 y 5 0

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 0 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 (1)    6 g e 9 (2)     6 9 g e (3)

After solving equations (2) & (3) we get: e   9 g  0 Then substitute in equation (1) we get: f  2 And the equation of the circle will be:     2 2 4 9 0 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 0 x y f x g y e Substitute with the three points (-1, 2), (-4,3) into the equation of the circle, we get the following two equations:      (1) 2 4 5 f g e      (2) 8 f 6 g e 25

Substitute with the center coordinates (-f,-g) into the equation of the line 4x-3y = 5, we get:    4 f 3 g 5 (3) After solving equations (1), (2) & (3) we get: f  7 g  11 e   35 And the equation of the circle will be:      2 2 x y 14 x 22 y 35 0