College Algebra, 2013a

500 Hooked on Conics
It is possible to obtain equations like (x − 3) 2 +(y +1) 2 = 0 or (x − 3) 2 +(y +1) 2 = −1, neither of
which describes a circle. (Do you see why not?) The reader is encouraged to think about what, if
any, points lie on the graphs of these two equations. The next example uses the Midpoint Formula,
Equation 1.2, in conjunction with the ideas presented so far in this section.
Example 7.2.4. Write the standard equation of the circle which has (−1, 3) and (2, 4) as the
endpoints of a diameter.
Solution. We recall that a diameter of a circle is a line segment containing the center and two
points on the circle. Plotting the given data yields
y
4
3
r
(h, k)
2
1
−2 −1 1 2 3
x
Since the given points are endpoints of a diameter, we know their midpoint (h, k) is the center of
the circle. Equation 1.2 gives us
(
x1 + x 2
(h, k) = , y )
1 + y 2
(
2 2
−1+2
= , 3+4 )
(
2 2
1
=
2 , 7 2)
The diameter of the circle is the distance between the given points, so we know that half of the
distance is the radius. Thus,
r = 1 √
(x 2 − x 1 ) 2 +(y 2 − y 1 ) 2
2
(√
10
Finally, since
2
= 1 2√
(2 − (−1)) 2 +(4− 3) 2
= 1 3
2√ 2 +1 2
√
10
=
2
) 2
= 10
(x
4 , our answer becomes − 1 2
) 2 (
+ y − 7 ) 2
= 10 2 4