Calculus: Analytic Geometry and Calculus, with Vectors

1.4 Distances, circles, and parabolas 25

and P2 lie on the same horizontal line and the formula reduces to the
correct formula d = 1x2 - x11 If x2 = x1, then Pi and P2 lie on the same
vertical line and the formula reduces to the correct formula d = Iy2 - y11.
When P1(x1,y1) and P2(x2,y2) are two given points for which X2 5'!5 x, and
y2 0 yi, we can, as in Figures 1.42 and 1.421, let Q(x2,yi) be the point on

P2(x2, y2) P2(x2, Y2)

Iy2-Y11 Iy2-y11

Q(x2,y1) Q(x2,y1)

Figure 1.42

1x2-x11 Pl(xl,y1)

Figure 1.421

the horizontal line through Pi and on the vertical line through P2. The

length of the horizontal line segment P1Q is x2 - x1 if x2 - xl > 0, is

x1 - x2 if x1 - x2 >= 0, and is 1x2 - x11 in each case. The length of the
vertical line segment QP2 is y2 - Yi if y2 - yi >- 0, is y1 - Y2 if
Y1 - y2 > 0, and is Iy2 - yll in each case. With the understanding that
the distance d between P1 and P2 is the length of the line segment PiP2,
we can therefore apply the Pythagoras theorem to the right triangle P1QP2
to obtain

d2 = 1x2 - x112 + Iy2 - Yi12

and hence

d2 = (x2 - x1)2 + (y2 - yi)2.

Since d >- 0, taking square roots gives the
required formula (1.41).
We are all familiar with the fact, illustrated

x

Figure 1.43

in Figure 1.43, that the circle C with center at Po(h,k) and radius a is the
set of points in the plane whose distances from Po are equal to the radius a.
From the distance formula, we see that the point P(x,y) lies on this circle
if and only if

(1.44) (x - h) l + (y- k)' =a2.

This is therefore the equation of the circle with center at (h,k) and radius
a. We must always remember this and the fact that

(1.45) x2 + y2 = a2

is the equation of the circle with center at the origin and radius a.
The equation of the circle with center at (-2,3) and radius 5 is

(1.451) (x + 2)2 + (y - 3)2 = 25.