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Distance and Midpoint Formulas; Circles

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Juan Jorrin

on 19 June 2017

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Transcript of Distance and Midpoint Formulas; Circles

Distance and Midpoint Formulas; Circles
Find the Midpoint of a Line Segment
Midpoint
Write the Standard Form of a
Circle’s Equation
Finding the Standard Form of a Circle’s Equation
Example:
Write the standard form of the equation of the circle with center (0,3) and radius 2. Graph the circle.
Example: Find the midpoint of the given line segment.
Using , identify the center and the radius of the circle.
The General Form of a Circle
Give the Center and Radius of a Circle Whose Equation is in Standard Form
Find the Distance Between Two Points
Distance Formula

used to find the distance between two points
A(x , y )
and
B(x , y )
1
1
2
2
d = √(x - x ) + (y - y )
1
1
2
2
2
2
Proof of the Distance Formula
Let
(x ,y )
denote the coordinates of
P
, and let
(x ,y )
denote the coordinates of
P
.
x
x
y
y
(x ,y )
(x ,y )
(x ,y )
1
1
2
2
1
1
2
1
2
2
|x - x |
|y - y |
2
1
2
1
1
1
1
2
2
2
By the Pythagorean Theorem
d = |x - x | + |y - y |
Since,
|x - x | = (x - x )

and

|y - y | = (y - y )
Therefore,
d = √(x - x ) + (y - y )
1
1
2
2
2
2
2
2
2
2
2
1
1
2
1
2
1
2
1
2
1
2
2
2
2
P
1
P
2
d
Example: Using the Distance Formula, find the distance between (-3,5) and (7,-1).
Solution:
Let
(
x
,
y
) = (-3,5)
and
(
x
,
y
) = (7,-1)
.
d = √(x - x ) + (y - y )
1
1
2
2
2
2
=√(7 - (-3)) + ((-1) - 5)
=√(7 + 3) + (-1 - 5)
=√(10) + (-6)
=√100 + 36
=√136
d=2√34
Thus, the distance between the given points is
2√34
or approximately
11.66
units.
2
2
2
2
2
2
2
2
1
1
the point halfway between the endpoints of a line segment
is the point on that line segment that divides the segment two congruent segments
Midpoint Formula
The midpoint of segment with endpoints (
x , y
) and (
x , y
) has coordinates
x + x
2
y + y
2
1
1
2
2
M
,
1
1
2
2
Solution:
Identify the given:
x
1
x
2
y
1
y
2
x = -3
1
y = 4
1
x = 2
2
y = 1
2
Use the midpoint formula:
Plug in the given values:
x + x
2
y + y
2
1
1
2
2
M
,
-3 + 2
2
4 + 1
2
M
,
Simplify the numerator:
-1
2
5
2
M
,
Circle
is a locus (set) of points in a plane equidistant from a fixed point called center
The fixed distance from the circle’s center to any point on the circle is called the radius.
Standard Form of the Equation of a Circle
(x - h) + (y - k) = r
with center
(h,k)
and radius
r
2
2
2
Solution:
center (0,3)
h = 0
k = 3
radius
r
= 2
(x -
0
) + (y -
3
) =
2
(x - h) + (y - k) = r
2
2
2
2
2
2
x + (y - 3) = 4
2
2
Example:
This is the given equation of a circle
(x - 5) + (y + 2) = 16
2
2
a. Find the center and radius .
Compare the given equation to the standard form of the equation of a circle.
(x - h) + (y - k) = r
2
2
2
(x - 5) + (y + 2) = 16
2
2
(x -
5
) + (y - (
- 2
)) =
4
2
2
2
h
k
r
Then, the circle has center at (5,-2) and radius 4.
b. Graph the equation.
Locate the center (5,-2).
(5,-2)
Because r =4, locate 4 points on the circle by going 4 units to the right, to the left, up and down from the center.
c. Use the graph to identify the relation's domain and range.
The four points on the circle can be used to determine the domain and range.
(5,2)
(9,2)
(5,-6)
(1,-2)
Domain
Range
Domain: [1,9]
Range: [2,-6]
Convert the General Form of a Circle’s
Equation to Standard Form
x + y + Dx + Ey + F = 0
2
2
where D, E and F are real numbers
Example: W
rite in standard form and graph.
Given:
x + y - 2x - 4y - 4
= 0
Solution: Group the x's and y's on the left side of the equation.
(x - 2x) + (y - 4y) = 4
Complete the square. Remember to add it to the right hand side, so the equation is balance.
(x - 2x + 1) + (y - 4y + 4) = 4 + 1 + 4
(x - 2x + 1) + (y - 4y + 4) = 9
Factor the left side.
(x - 1) + (y - 2) = 3
2
2
2
2
2
2
2
2
2
2
2
(x - 1) + (y - 2) = 3
2
2
2
C
(1,2)

r
= 3

Then, the graph of is
(x - 1) + (y - 2) = 3
2
2
2
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