Quadrant II where x is negative and y is positive You should be familiar with the rectangular coordinate system and point plotting from an earlier algebra course. Let's just run through the basics. y axis Quadrant II where x is negative and y is positive Quadrant I where both x and y are positive x axis origin Quadrant IV where x is positive and y is negative Quadrant III where both x and y are negative

Let's plot the point (-3,-5) 8 7 6 5 4 3 2 1 2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 4 6 8 -2 -3 -4 -5 -6 -7 Let's plot the point (0,7) Let's plot the point (-3,-5)

So the distance from (-6,4) to (1,4) is 7. We now want to find the distance between two points. 7 units apart 8 7 6 So the distance from (-6,4) to (1,4) is 7. 5 4 (-6,4) (1,4) 3 2 1 2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 4 6 8 -2 -3 -4 -5 -6 -7 If the points are located horizontally from each other, the y coordinates will be the same. You can look to see how far apart the x coordinates are.

So the distance from (-6,4) to (-6,-3) is 7. What coordinate will be the same if the points are located vertically from each other? 8 7 6 5 4 (-6,4) 3 2 7 units apart 1 2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 4 6 8 -2 -3 So the distance from (-6,4) to (-6,-3) is 7. (-6,-3) -4 -5 -6 -7 If the points are located vertically from each other, the x coordinates will be the same. You can look to see how far apart the y coordinates are.

The Pythagorean Theorem will help us find the hypotenuse But what are we going to do if the points are not located either horizontally or vertically to find the distance between them? 8 7 Let's start by finding the distance from (0,0) to (4,3) 6 5 4 5 3 ? 2 3 1 2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 4 6 8 -2 4 -3 The Pythagorean Theorem will help us find the hypotenuse -4 -5 -6 So the distance between (0,0) and (4,3) is 5 units. -7 This triangle measures 4 units by 3 units on the sides. If we find the hypotenuse, we'll have the distance from (0,0) to (4,3) Let's add some lines and make a right triangle.

Again the Pythagorean Theorem will help us find the hypotenuse Now let's generalize this method to come up with a formula so we don't have to make a graph and triangle every time. 8 Let's start by finding the distance from (x1,y1) to (x2,y2) 7 (x2,y2) 6 5 ? 4 y2 – y1 3 (x1,y1) 2 1 2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 4 6 8 x2 - x1 -2 -3 Again the Pythagorean Theorem will help us find the hypotenuse -4 -5 -6 -7 Solving for c gives us: Let's add some lines and make a right triangle. This is called the distance formula

3 -5 Let's use it to find the distance between (3, -5) and (-1,4) Plug these values in the distance formula (x1,y1) (x2,y2) means approximately equal to -1 3 4 -5 CAUTION! found with a calculator Don't forget the order of operations! You must do the parenthesis first then powers (square the numbers) and then add together BEFORE you can square root

First let’s look at the official formula for finding the midpoint. If we have two points and we want to find the point halfway between them, it is called the midpoint. Let's start by finding the midpoint between (0,0) and (4,3) 8 7 6 5 (4, 3) 4 3 2 1 2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 4 6 8 (0, 0) -2 -3 First let’s look at the official formula for finding the midpoint. -4 -5 -6 -7 All this is, is an average of the x’s from the endpoints to get the x value of the midpoint and an average of the y’s from the endpoints to get the y value of the midpoint. This is easier to remember than memorizing a formula.

Absolute value is the distance from zero Absolute value is the distance from zero. It doesn't matter whether we are in the positive direction or the negative direction, we just care about how far away we are. 4 units away from 0 4 units away from 0 2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 4 6 8

If we want to know how far apart points on the number line are, we can take the difference between them and then take the absolute value. What is the distance from -5 to 3? 8 units apart 2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 4 6 8

CIRCLES © 2002 by Shawna Haider

The standard form of the equation of a circle with its center at the origin is r is the radius of the circle so if we take the square root of the right hand side, we'll know how big the radius is. Notice that both the x and y terms are squared. When we looked at parabolas, only the x term was squared.

Let's look at the equation This is r2 so r = 3 The center of the circle is at the origin and the radius is 3. Let's graph this circle. 2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 4 6 8 Count out 3 in all directions since that is the radius Center at (0, 0)

The center of the circle is at (h, k). This is r2 so r = 4 If the center of the circle is NOT at the origin then the equation for the standard form of a circle looks like this: The center of the circle is at (h, k). This is r2 so r = 4 Find the center and radius and graph this circle. The center of the circle is at (h, k) which is (3,1). 2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 4 6 8 The radius is 4

If you take the equation of a circle in standard form for example: This is r2 so r = 2 (x - (-2)) Remember center is at (h, k) with (x - h) and (y - k) since the x is plus something and not minus, (x + 2) can be written as (x - (-2)) You can find the center and radius easily. The center is at (-2, 4) and the radius is 2. But what if it was not in standard form but multiplied out (FOILED) Moving everything to one side in descending order and combining like terms we'd have:

If we'd have started with it like this, we'd have to complete the square on both the x's and y's to get in standard form. Move constant to the other side Group x terms and a place to complete the square Group y terms and a place to complete the square 4 16 4 16 Complete the square Write factored and wahlah! back in standard form.

If we'd have started with it like this, we'd have to complete the square on both the x's and y's to get in standard form. Move constant to the other side Group x terms and a place to complete the square Group y terms and a place to complete the square 4 16 4 16 Complete the square Write factored and wahlah! back in standard form.

7 Now let's work some examples: Find an equation of the circle with center at (0, 0) and radius 7. Let's sub in center and radius values in the standard form 7

Find an equation of the circle with center at (0, 0) that passes through the point (-1, -4). Since the center is at (0, 0) we'll have The point (-1, -4) is on the circle so should work when we plug it in the equation: Subbing this in for r2 we have:

Find an equation of the circle with center at (-2, 5) and radius 6 Subbing in the values in standard form we have: -2 5 6

Find an equation of the circle with center at (8, 2) and passes through the point (8, 0). Subbing in the center values in standard form we have: 8 2 Since it passes through the point (8, 0) we can plug this point in for x and y to find r2.