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Appendix C C-81
The point P also lies on the circle with center C and radius r, so the distance
from C to P is r. Using vectors, the length of the vector P C is r, that is
P C r (3)
We can solve the system of equations (2) and (3) for t by substituting the expression
for P from equation (2) into equation (3) and rearranging to get
tA (P 0 C) r (4)
Squaring both sides of equation (4) we get
tA (P 0 C) 2 r 2 (5)
Now (from Exercise 64 in Section 10.4)* the squared magnitude of a vector is
the dot product of the vector with itself. So
[tA (P 0 C)] # [tA (P 0 C)] r 2 (6)
Then [using Exercises 61(c) and 62(d) in Section 10.4] we have
tA # [tA (P 0 C)] (P 0 C) # [tA (P 0 C)] r 2 (7)
Exercise 3
to obtain
#
Expand the left-hand side of equation (7) and regroup the terms
(tA) (tA) (tA) (P 0 C) (P 0 C) (tA) (P 0 C) (P 0 C) r 2 (8)
Exercise 4
(a) Show that if A and B are any two vectors and s is a real number, then (sA) # B
s(A#
B) A # (sB). Hint: Let A 8x 1 , y 1 9 and B 8x 2 , y 2 9.
(b) Use part (a) to put equation (8) in the form of a quadratic equation in t to
obtain
(A#
A)t 2 [2A # (P 0 C)]t [(P 0 C) # (P 0 C) r 2 ] 0 (9)
Then show that t satisfies the quadratic equation
A 2 t 2 [2A # (P 0 C)]t ( P 0 C 2 r 2 ) 0 (10)
Note that the coefficients in equation (9) are written completely in terms of dot
products, which is most convenient for computer computations. In equation (10)
two of the coefficients look a little simpler in terms of lengths of vectors.
Exercise 5
to obtain
#
Use the quadratic formula to solve equation (10) for t and simplify
t A # (P0 C) 2[A # (P0 C)] 2 0 A 0
2 ( 0 P 0 C 0
2 r 2 )
2
0 A 0
#
Then explain the geometric significance of the numerical value of the expression
[A # (P 0 C)] 2 A 2 ( P 0 C 2 r 2 ) being positive, negative, or zero.
#
(11)
*Precalculus: A Problems-Oriented Approach, 7th edition