5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book May 8, 2018 13:46

Review of Precalculus 49

y

0 x

y = x + 1

(1, 2)

A (3, 0)

B (−1, 4)

Figure 5.1-6

Begin by finding the midpoint ofAB. Midpoint=

(

3 +(−1)

2

,

0 + 4

2

)

=(1, 2). The slope

ofABism =

4 − 0

− 1 − 3

=−1. Therefore, the perpendicular bisector of ABhas a slope of

1. Since the perpendicular bisector ofABpasses through the midpoint, you could use the
   point-slope form to obtainy− 2 =1(x−1) ory− 2 =x−1ory =x+1.

Example 3

Write an equation of the circle with center atC(−2, 1) and tangent to linelhaving the
equationx+y=5. (See Figure 5.1-7.)

y

0 x

x + y = 5

M

C (−2, 1)

Figure 5.1-7

LetMbe the point of tangency. Express the equationx+y=5 in slope-intercept form to
obtainy=−x+5. Thus, the slope of linelis−1. SinceCMis a radius drawn to the point
of tangency, it is perpendicular to linel, and the slope ofCMis 1. Using the point-slope
formula, the equation ofCMisy− 1 =1(x−(−2)) ory=x+3. To find the coordinates
of pointM, solve the two equations y=−x+5 andy =x+3 simultaneously. Thus,
−x+ 5 =x+3 which is equivalent to 2= 2 xorx=1.

Substitutingx=1 intoy=x+3, you havey=4. Therefore, the coordinates ofMare
(1, 4). SinceCMis the radius of the circle, you should find the length ofCMby using the
distance formulad=

√

(x 2 −x 1 )^2 +(y 2 −y 1 )^2. Thus,CM=

√

(1−(−2))^2 +(4−1)^2 =

√
18.
Now that you know both the radius of the circle (r=

√
18) and its center, (−2, 1), use the
formula (x−h)^2 +(y−k)^2 =r^2 to find an equation of the circle. Thus, an equation of the
circle is (x−(−2))^2 +(y−1)^2 =18 or (x+2)^2 +(y−1)^2 =18.