Distance, Midpoints, and Circles - Battaly · 2.8 Distance, Midpoints, and Circles Class Notes:...

Distance, Midpoints, and Circles

© G. Battaly 2015 1

November 22, 2015

2.8 Distance, Midpoints, and Circles

Class Notes: Prof. G. Battaly, Westchester Community College, NY

Study 2.8 CVC # 15; # 119; 23, 27, 31, 35, 39 41, 45, 49, 53, 57, 61

College Algebra & Trig Home Page Homework

GOALS:1. Understand the distance formula.2. Find the distance between 2 points.3. Find the midpoints of the line segment joining 2 points.4. Recognize the general and standard forms of the equation for a circle.5. Identify the center and radius of a circle from the equation in standard form.6. Find the center and radius of a circle by converting from general form to standard form.



Pythagorean Theorum

a2 + b2 = c2


ac

b

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November 22, 2015



Pythagorean Theorum

a2 + b2 = c2


ac

b

To find c, knowing a and b, need to solve for c:

c2 = a2 + b2

c = √ a2 + b2



Pythagorean Theorum

a2 + b2 = c2


ac

b

(x1,y1)

(x2,y2)

To find c, knowing a and b, need to solve for c:

c2 = a2 + b2

c = √ a2 + b2 b = x2 x1

a = y2 y1then:c = √ (x2 x1)2 + (y2 y1)2

Distance from (x1,y1) to (x2,y2)

d = √ (x2 x1)2 + (y2 y1)2




November 22, 2015





d = √ (x2 x1)2 + (y2 y1)2

G: (5,1) and (8,5) F: distance between pts.

2 4 6 8

8

6

4

2

d = √ (x2 x1)2 + (y2 y1)2

d = √ ( )2 + ( )2

2 4 6 8

8

6

4

2





d = √ (x2 x1)2 + (y2 y1)2


d = √ (x2 x1)2 + (y2 y1)2

d = √ ( )2 + ( )2

d = √ ( 8 5)2 + ( 5 1)2

d = √ ( 3)2 + ( 4)2

= √ 25 = 5




November 22, 2015





d = √ (x2 x1)2 + (y2 y1)2

d = √ ( )2 + ( )2

d = √ (x2 x1)2 + (y2 y1)2

(4,1)

(2,3)

d





d = √ (x2 x1)2 + (y2 y1)2

d = √ ( )2 + ( )2

d = √ [2 (4)]2 + [3 (1)]2

d = √ ( 6)2 + (2)2

= √ 40 = 2√ 10

d = √ (x2 x1)2 + (y2 y1)2

(4,1)

(2,3)

d




November 22, 2015

ac

b(x1,y1)

(x2,y2)




Find the midpoint of line b:

half the distance = x2 x1 2add to x1: x2 x1 + 2x1 2 2midpoint at x = x1 + x2 2

ac

b(x1,y1)

(x2,y2)




Find the midpoint of line a: half the distance = y2 y1 2add to y1: y2 y1 + 2y1 2 2midpoint at y = y1 + y2 2




November 22, 2015

ac

b(x1,y1)

(x2,y2)




Find the midpoint of line c: midpoint at:

x1 + x2 , y1 + y2 2 2




G: (10,4) and (2,6) F: midpoint of line segment joining pts.

x1 + x2 , y1 + y2 2 2

(6,5)

x1 + x2 , y1 + y2 2 2

(10,4)

(2,6)




November 22, 2015





x1 + x2 , y1 + y2 2 2

(6,5)

x1 + x2 , y1 + y2 2 2

10 + 2 , 4 + 6 2 2

12 , 10 2 2

6 , 5

(2,6)

(10,4)





x1 + x2 , y1 + y2 2 2

(2,1)

(8,6)

x1 + x2 , y1 + y2 2 2




November 22, 2015





x1 + x2 , y1 + y2 2 2

(2,1)

(8,6)

x1 + x2 , y1 + y2 2 2

(2) + (8) , (1) + 6 2 2

10 , 5 2 2

5 , 5 2




Circle: Set of all Points Equidistant from a fixed point. The fixed point is the Center. The fixed distance is the Radius

Cr




November 22, 2015




Circle: Standard Form: (xh)2 + ( yk)2 = r2

C (h,k)r (x,y)




Circle: General Form: x2 + y2 + Dx + Ey + F = 0

C (h,k)r (x,y)




November 22, 2015




Circle: x2 + y2 = 25

(xh)2 + ( yk)2 = r2

∴ C(0,0) and r = 5




Circle: x2 + y2 = 25

(xh)2 + ( yk)2 = r2

∴ C(0,0) and r = 5

C(0,0)

r = 5




November 22, 2015




(xh)2 + ( yk)2 = r2

∴ C( __, __ ) and r = __

G: (x2)2 + (y1)2 = 16 F: C, r, sketch




(xh)2 + ( yk)2 = r2

∴ C(2,1) and r = 4

C(2,1)

r = 4

G: (x2)2 + (y1)2 = 16 F: C, r, sketch




November 22, 2015




G: x2 + y2 + 4x + 2y + 1 = 0 F: C, r, sketch

x2 + y2 + Dx + Ey + F = 0

x2 + 4x + __ + y2 + 2y + __ = 1 + __




x2 + y2 + Dx + Ey + F = 0

x2 + 4x + _4_ + y2 + 2y + _1_ = 1 + _5_

(x + _)2+ (y + _)2 = _

G: x2 + y2 + 4x + 2y + 1 = 0 F: C, r, sketch

∴ C(__,__), r = __




November 22, 2015




x2 + y2 + Dx + Ey + F = 0

x2 + 4x + _4_ + y2 + 2y + _1_ = 1 + _5_

(x + 2)2+ (y + 1)2 = 4

G: x2 + y2 + 4x + 2y + 1 = 0 F: C, r, sketch

∴ C(2,1), r = 2

C(2,1)r = 2

2

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