1. Find: R [ft] y [ft] C (5,6) B (2,5) 4.6 5.1 5.8 6.4 A (0,0) x [ft]
Find the radius of the curve, in feet. [pause] In this problem, points A, B and C ----
A (0,0)
x [ft]

2. Find: R [ft] y [ft] C (5,6) B (2,5) 4.6 5.1 5.8 6.4 A (0,0) x [ft]
are plotted on a x y plane, where the coordinates ---
A (0,0)
x [ft]

3. Find: R [ft] y [ft] C (5,6) B (2,5) 4.6 5.1 5.8 6.4 A (0,0) x [ft]
of all 3 points are provided. [pause] If a circular curve is drawn ---
A (0,0)
x [ft]

4. Find: R [ft] y [ft] C (5,6) B (2,5) 4.6 5.1 5.8 6.4 A (0,0) x [ft]
connecting these three points, the problem asks, what is the radius ---
A (0,0)
x [ft]

5. Find: R [ft] C (5,6) y [ft] B (2,5) O (Ox,Oy) R A (0,0) x [ft]
of a such a curve. We’ll equate the radius to the distance between points ----
A (0,0)
x [ft]

6. Find: R [ft] C (5,6) y [ft] B (2,5) R= (Ox-Ax)2+(Oy-Ay)2 O (Ox,Oy) R
A and O. Since point A is at the origin, the equation for the radius ----
A (0,0)
x [ft]

7. Find: R [ft] C (5,6) y [ft] B (2,5) R= (Ox-Ax)2+(Oy-Ay)2 R= Ox2 + Oy2
O (Ox,Oy) R
simplifies to the root sum squared of the x and y coordinates of point O, ---
A (0,0)
x [ft]

8. Find: R [ft] C (5,6) y [ft] B (2,5) R= (Ox-Ax)2+(Oy-Ay)2 R= Ox2 + Oy2
O (Ox,Oy) R
which is a point defined at the center of the curve. We’ll find the coordinates of Point O ---
A (0,0)
x [ft]

9. Find: R [ft] R= Ox2 + Oy2 C (5,6) y [ft] B (2,5) O (Ox,Oy) R A (0,0)
by finding the intersection of lines, D O, and ---
A (0,0)
x [ft]

10. Find: R [ft] R= Ox2 + Oy2 C (5,6) y [ft] B (2,5) E D O (Ox,Oy) R
and E O. We can define Line D O by ----
A (0,0) x

11. Find: R [ft] R= Ox2 + Oy2 C (5,6) y [ft] B (2,5) E D O (Ox,Oy) R
drawing Chord A B, and then drawing a line normal to Chord A B which ----
A (0,0) x

12. Find: R [ft] R= Ox2 + Oy2 C (5,6) y [ft] B (2,5) E D O (Ox,Oy) R
passes through the midpoint of Chord A B, which occurs at Point D. Line E O is generated in a similar fashion, ----
A (0,0) x

13. Find: R [ft] R= Ox2 + Oy2 C (5,6) y [ft] B (2,5) E D O (Ox,Oy) R
Chord B C is drawn out, and line E O ---
A (0,0) x

14. Find: R [ft] R= Ox2 + Oy2 C (5,6) y [ft] B (2,5) E D O (Ox,Oy) R
bisects and is perpendicular to Chord B C. Using geometry, ---
A (0,0) x

15. Find: R [ft] R= Ox2 + Oy2 C (5,6) y [ft] B (2,5) E D O (Ox,Oy) R
we can show how the center of the curve, Point O, must coincide at ---
A (0,0) x

16. Find: R [ft] R= Ox2 + Oy2 C (5,6) y [ft] B (2,5) E D O (Ox,Oy) R
the intersection of lines D O and E O. Both Triangle A B O and ----
A (0,0) x

17. Find: R [ft] R= Ox2 + Oy2 C (5,6) y [ft] B (2,5) E D O (Ox,Oy) A (0,0)
Triangle B C O are isosceles, since line segments A O, B O and ---
A (0,0) x

18. Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) E R R D x R O (Ox,Oy)
C O all equal to the radius of the curve. Then, from the law of lines, --- R
O (Ox,Oy)
A (0,0)

19. Find: R [ft] = = R= Ox2 + Oy2 C (5,6) y B (2,5) sin(ABAO) sin(AABO) E
LBO
LAO R R D
sin(ACBO)
sin(ABCO) =
LCO
LBO x
if we substitute in the radius, for the lengths A O, B O and C O, --- R
O (Ox,Oy)
A (0,0)

20. Find: R [ft] = = R= Ox2 + Oy2 C (5,6) y B (2,5) sin(ABAO) sin(AABO) E
LBO
LAO R R D
sin(ACBO)
sin(ABCO) =
LCO
LBO x
the denominators and sine terms drop out, such that, Angle B A O equals --- R
O (Ox,Oy)
A (0,0)

21. Find: R [ft] = = R= Ox2 + Oy2 C (5,6) y B (2,5) sin(ABAO) sin(AABO) E
LBO
LAO R R
ABAO=AABO D
sin(ACBO)
sin(ABCO) =
LCO
LBO x
Angle A B O, and Angle C B O equals Angle B C O. For these 2 pairs of angles to be equal, ---
ACBO=ABCO R
O (Ox,Oy)
A (0,0)

22. Find: R [ft] = = R= Ox2 + Oy2 C (5,6) y B (2,5) sin(ABAO) sin(AABO) E
LBO
LAO
ABAO=AABO D
sin(ACBO)
sin(ABCO) =
LCO
LBO x
Point O, must lie on the lines which bisect, and are perpendicular to, Chords A B and B C. If Point O lies not on either Line D O or Line E O, --- O
ACBO=ABCO
A (0,0)

23. Find: R [ft] = = R= Ox2 + Oy2 C (5,6) y B (2,5) sin(ABAO) sin(AABO) E
LBO
LAO
ABAO=AABO D
sin(ACBO)
sin(ABCO) =
LCO
LBO x
this would violate the angle equalities, which we know ---
ACBO=ABCO
A (0,0) O

24. Find: R [ft] = = R= Ox2 + Oy2 C (5,6) y B (2,5) sin(ABAO) sin(AABO) E
LBO
LAO
ABAO=AABO D
sin(ACBO)
sin(ABCO) =
LCO
LBO x
to be true based on geometry. [pause] Therefore, all we have to do is ---
ACBO=ABCO
A (0,0) O

25. Find: R [ft] = = R= Ox2 + Oy2 C (5,6) y B (2,5) sin(ABAO) sin(AABO) E
LBO
LAO
ABAO=AABO D
sin(ACBO)
sin(ABCO) =
LCO
LBO x
solve for the point where lines D O and E O intersect. The general equation for a line in 2 dimensions is --- O
ACBO=ABCO
A (0,0)
line DO
line EO

26. Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) y-yo = m * (x-xo) E D x O
y minus y not equals m times the quantity x minus x not. In this equation, x not and y not --- O
A (0,0)
line DO
line EO

27. Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) y-yo = m * (x-xo) E
known coordinates D
of a point on the line x
are known coordinates of a point on the line. So for line D O, we’ll determine ---- O
line DO
A (0,0)
line EO
y-yo = m * (x-xo)
y-yo = m * (x-xo)

28. Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) y-yo = m * (x-xo) E
known coordinates D
of a point on the line x
the x and y coordinates of Point D, and then substitute those values in for ---- O
line DO
A (0,0)
line EO
y-yo = m * (x-xo)
y-yo = m * (x-xo)

29. Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) y-yo = m * (x-xo) E
known coordinates D
of a point on the line x
x not and y not. For line E O, we’ll use the x and y coordinates for point E --- O
A (0,0)
line DO
line EO
y-Dy = m * (x-Dx)
y-yo = m * (x-xo)

30. Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) y-yo = m * (x-xo) E
known coordinates D
of a point on the line x
[pause] The variable m represents the slope of the line, --- O
A (0,0)
line DO
line EO
y-Dy = m * (x-Dx)
y-Ey = m * (x-Ex)

31. Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) y-yo = m * (x-xo) E slope
known coordinates D
of a point on the line x
in y over x. To the equations for line D O and line E O, we’ll add a subscript --- O
A (0,0)
line DO
line EO
y-Dy = m * (x-Dx)
y-Ey = m * (x-Ex)

32. Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) y-yo = m * (x-xo) E slope
known coordinates D
of a point on the line x
D O and E O, to their slope terms. [pause] Next we’ll find the coordinates of Points D and E, --- O
A (0,0)
line DO
line EO
y-Dy = mDO * (x-Dx)
y-Ey = mEO * (x-Ex)

33. Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) Ax+Bx Ay+By Dx= Dy= E 2 2
Bx+Cx
By+Cy D
Ex=
Ey= 2 2
by averaging the given coordinates of Points A and B, and Points B and C, respectively. O
A (0,0)
line DO
line EO
y-Dy = mDO * (x-Dx)
y-Ey = mEO * (x-Ex)

34. Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) Ax+Bx Ay+By Dx= Dy= E 2 2
Bx+Cx
By+Cy D
Ex=
Ey= 2 2
After plugging in the coordinates, we know point D is located at --- O
A (0,0)
line DO
line EO
y-Dy = mDO * (x-Dx)
y-Ey = mEO * (x-Ex)

35. Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) 0+2 0+5 Dx= Dy= E 2 2 Dx=1
2+5
5+6 D
Ex=
Ey= 2 2
x equals 1, y equals 2.5 and Point E is located at x equals 3.5, y equals After plugging in these coordinates, ---
Ex=3.5
Ey=5.5 O
A (0,0)
line DO
line EO
y-Dy = mDO * (x-Dx)
y-Ey = mEO * (x-Ex)

36. Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) 0+2 0+5 Dx= Dy= E 2 2 Dx=1
2+5
5+6 D
Ex=
Ey= 2 2
we still have to solve for the slopes of lines D O and ---
Ex=3.5
Ey=5.5 O
A (0,0)
line DO
line EO
y-2.5 = mDO * (x-1)
y-5.5= mEO * (x-3.5)

37. ? ? Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) 0+2 0+5 Dx= Dy= E 2 2
2+5
5+6 D
Ex=
Ey= 2 2
E O. First we’ll find the slope of lines A B and ----
Ex=3.5
Ey=5.5 O
A (0,0) ?
line DO ?
line EO
y-2.5 = mDO * (x-1)
y-5.5= mEO * (x-3.5)

38. ? ? Find: R [ft] R= Ox2 + Oy2 C (5,6) y (By-Ay) B (2,5) mAB = E
(Bx-Ax)
(Cy-By)
mBC =
(Cx-Bx) D
B C. Then, we’ll determine the slopes of lines D O and E O since the slopes of ---- O
A (0,0) ? ?
y-2.5 = mDO * (x-1)
y-5.5= mEO * (x-3.5)

39. Find: R [ft] R= Ox2 + Oy2 C (5,6) y (By-Ay) B (2,5) mAB = E (Bx-Ax)
mBC =
(Cx-Bx) D
mDO =-mAB-1
perpendicular lines in 2 dimensions are negative reciprocals of each other. After substituting in --- O
mEO =-mBC-1
A (0,0)
y-2.5 = mDO * (x-1)
y-5.5= mEO * (x-3.5)

40. Find: R [ft] R= Ox2 + Oy2 C (5,6) y (By-Ay) B (2,5) mAB = E (Bx-Ax)
mBC =
(Cx-Bx) D
mDO =-mAB-1
the coordinates of Points A, B and C, we find the slope of line A B equals 2.5, --- O
mEO =-mBC-1
A (0,0)
y-2.5 = mDO * (x-1)
y-5.5= mEO * (x-3.5)

41. Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) (5-0) mAB = =2.5 E (2-0)
(6-5)
mBC =
=0.333
(5-2) D
mDO =-mAB-1
and the slope of Line B C equals, After taking the negative reciprocal of these values --- O
mEO =-mBC-1
A (0,0)
y-2.5 = mDO * (x-1)
y-5.5= mEO * (x-3.5)

42. Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) (5-0) mAB = =2.5 E (2-0)
(6-5)
mBC =
=0.333
(5-2) D
mDO =-mAB-1
we find the slopes of lines D O and E O, which equal, --- O
mEO =-mBC-1
A (0,0)
y-2.5 = mDO * (x-1)
y-5.5= mEO * (x-3.5)

43. Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) (5-0) mAB = =2.5 E (2-0)
(6-5)
mBC =
=0.333
(5-2) D
mDO =-mAB-1
=-0.4
negative 0.4 and negative 3. Now we have the equations for line D O and line E O, which simplify to --- O
mEO =-mBC-1
=-3
A (0,0)
y-2.5 = mDO * (x-1)
y-5.5= mEO * (x-3.5)

44. Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) 0.4 * x + y = 2.9 E
mDO =-2.5-1
=-0.4
0.4 times x plus y equals 2.9, and 3 times x plus y equals 16. Subtracting the first equation from the second, --- O
mEO =
=-3
A (0,0)
y-2.5 = -0.4 * (x-1)
y-5.5= -3 * (x-3.5)

45. Find: R [ft] +( ) R= Ox2 + Oy2 C (5,6) y B (2,5) -1*( )
-1*( )
0.4 * x + y = E
+( )
3 * x + y = 16.0
2.6 * x = 13.1 D
x = 5.04
we find x equals Substituting 5.04 in for x, we can solve for y, which equals --- O
A (0,0)
y-2.5 = -0.4 * (x-1)
y-5.5= -3 * (x-3.5)

46. Find: R [ft] +( ) R= Ox2 + Oy2 C (5,6) y B (2,5) -1*( )
-1*( )
0.4 * x + y = E
+( )
3 * x + y = 16.0
2.6 * x = 13.1 D
x = 5.04
and 0.88 correspond to the x and y coordinates of Point O, --- O
A (0,0)
y=0.88
y-2.5 = -0.4 * (x-1)
y-5.5= -3 * (x-3.5)

47. Find: R [ft] +( ) R= Ox2 + Oy2 C (5,6) y B (2,5) -1*( )
-1*( )
0.4 * x + y = E
+( )
3 * x + y = 16.0
2.6 * x = 13.1 D
x = 5.04
the center of the curve. Plugging these values into our equation ---
O (5.04,0.88)
A (0,0)
y=0.88
y-2.5 = -0.4 * (x-1)
y-5.5= -3 * (x-3.5)

48. Find: R [ft] +( ) R= Ox2 + Oy2 C (5,6) y B (2,5) -1*( )
-1*( )
0.4 * x + y = E
+( )
3 * x + y = 16.0
2.6 * x = 13.1 D
x = 5.04
for the radius, the radius of the curve equals, ---
O (5.04,0.88)
A (0,0)
y=0.88
y-2.5 = -0.4 * (x-1)
y-5.5= -3 * (x-3.5)

49. Find: R [ft] +( ) R= Ox2 + Oy2 C (5,6) y R= 5.12 [ft] B (2,5) -1*( )
-1*( )
0.4 * x + y = E
+( )
3 * x + y = 16.0
2.6 * x = 13.1 D
x = 5.04
5.12 feet. [pause]
O (5.04,0.88)
A (0,0)
y=0.88
y-2.5 = -0.4 * (x-1)
y-5.5= -3 * (x-3.5)

50. Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) R= 5.12 [ft] E 4.6 5.1 5.8
6.4 D
When reviewing the possible solutions, ---
O (5.04,0.88)
A (0,0)
y-2.5 = -0.4 * (x-1)
y-5.5= -3 * (x-3.5)

51. Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) R= 5.12 [ft] E AnswerB
4.6
5.1
5.8
6.4 D
the answer is B.
O (5.04,0.88)
A (0,0)
y-2.5 = -0.4 * (x-1)
y-5.5= -3 * (x-3.5)

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