1. COORDINATES, GRAPHS AND LINES
RAFIZAH KECHIL, UiTM PULAU PINANG
(Source: Cengage Learning, Hasfazilah Ahmat)

2. Plane Analytic Geometry

3. Circles

4. What You Should Learn Find equations of circles.
Sketch graphs of circles.

5. Circles
Consider the circle shown in Figure A point (x, y) is on the circle if and only if its distance from the center (h, k) is r.
By the Distance Formula,
Figure 1.12

6. Circles
The standard form of the equation of a circle with its center at the origin, (h, k) = (0, 0),
is simply
x2 + y2 = r 2.
Circle with center at origin

7. Example 8 – Finding the Equation of a Circle
The point (3, 4) lies on a circle whose center is at (–1, 2), as shown in Figure Write the standard form of the equation of this circle.
Figure 1.13

8. Example 8 – Solution
The radius of the circle is the distance between (–1, 2) and (3, 4).
Distance Formula
Substitute for x, y, h, and k.
Simplify.
Simplify.
Radius

9. Example 8 – Solution
cont’d
Using (h, k) = (–1, 2) and r = the equation of the circle is
(x – h)2 + (y – k)2 = r 2
[x – (–1)]2 + (y – 2)2 = ( )2
(x + 1)2 + (y – 2)2 = 20.
Equation of circle
Substitute for h, k, and r.
Standard form

11. Exercise: Question 64 Question 74 Question 16 Question 20 Section 1.1

12. Parabolas

13. Parabolas The graph of the quadratic function f (x) = ax2 + bx + c
is a parabola that opens upward or downward.
Parabola
Figure 4.20

14. Parabolas : Vertex (0,0) (a) Parabola with vertical axis
(b) Parabola with horizontal axis

15. Example 1:

16. Example 2 – A Parabola with a Horizontal Axis
Find the standard form of the equation of the parabola with vertex at the origin and focus at (2, 0).
Solution:
The axis of the parabola is horizontal,
passing through (0, 0) and (2, 0), as
shown in Figure 4.23.
Figure 4.23

17. Example 2 – Solution The standard form is y2 = 4px.
cont’d
The standard form is
y2 = 4px.
Because the focus is p = 2 units from the vertex, the equation is
y2 = 4(2)x
y2 = 8x.

18. Parabolas : Vertex=(h,k)
cont’d

19. Example 3 – Finding the Standard Equation of a Parabola
Find the vertex and focus of the parabola
x2 – 2x + 4y – 3 = 0.
Solution:
Complete the square to write the equation in standard form.
x2 – 2x + 4y – 3 = 0
x2 – 2x = –4y + 3
x2 – 2x + 1 = –4y
(x – 1)2 = –4y + 4
Write original equation.
Group terms.
Add 1 to each side.
Write in completed square form.

20. Example 3 – Solution (x – 1)2 = 4(–1)(y – 1)
cont’d
(x – 1)2 = 4(–1)(y – 1)
From this standard form, it follows that h = 1, k = 1, and p = –1. Because the axis is vertical and p is negative, the parabola opens downward.
The vertex is (h, k) = (1, 1) and the focus is (h, k + p) = (1, 0). (See Figure 4.42.)
Write in standard form, (x – h)2 = 4p(y – k).
Figure 4.42

21. Exercise:
Section 4.3
Question 22,26
Question 28,38