Multivariate Calculus - Bruce E. Shapiro

LECTURE 1. CARTESIAN COORDINATES 7
Example 1.8 Find the equation of sphere that is tangent to the plane x + y = 4
with center at the point P = (1, 1, 4). (See the text, page 599 #28.)
In two dimensions (the xy-plane), the equation x+y = 4 describes a line of slope
−1 with y-intercept of 4. This line also crosses the x-axis (has x-intercept) at x = 4
(see figure 1.4.) In three dimensions, this equation x + y = 4 represents a vertical
plane whose projection onto the xy-plane is the line x + y = 4.
To find the equation of the desired sphere, consider the horizontal plane passing
through P, namely, the plane z = 4. This plane sits parallel to and directly above
the xy-plane, and from an illustrative point of view (see figure 1.4) looks just like
the xy-plane. The center of the sphere will lay at (1, 1, 4), or at the point (1,1)
of the equivalent xy-plane look-alike illustration. The cross section of the sphere
with this plane is a circle of radius r (whose value is not yet determined) that is
tangent to the line x + y = 4 in this plane. The point of tangency is point on the
line that is closest to the point (1,1). By symmetry, this must be the midpoint of
the x-intercept and y-intercept, which is
( 0 + 4
2 , 4 + 0
2 , 4 + 4 )
= (2, 2, 4)
2
The radius is the distance between the center at (1,1,4) and the point of tangency
at (2,2,4):
r = √ (1 − 2) 2 + (1 − 2) 2 + (4 − 4) 2 = √ 2
Hence the equation of the sphere is
(x − 1) 2 + (y − 1) 2 + (z − 4) 2 = 2.
Math 250, Fall 2006 Revised December 6, 2006.