LetP1 =(x1, y1)andP2=(x2, y2)be two points in thexy-plane as shown in Fig. 1.
d
P1⫽ (x1, y1)
P2⫽ (x2, y2)
x2⫺ x1
y2⫺ y1 y
x
3 2 1
⫺2
⫺1
⫺1
⫺2
⫺4 ⫺3 1 2 3 4 5 6 P1⫽ (⫺4,3)
P2⫽ (5,⫺1) y
x
Figure 1 Figure 2
By Pythagoras’s theorem, stated in the appendix, the distancedbetweenP1andP2satisfies the equationd2=(x2−x1)2+(y2−y1)2. This gives the following important formula:
1 Of course, Fig. 8 is an idealization. The true income tax function is defined only for integral numbers of dollars—or, more precisely, it is a discontinuous “step function” which jumps up slightly whenever income rises by another dollar.
S E C T I O N 5 . 5 / D I S T A N C E I N T H E P L A N E . C I R C L E S 147 D I S T A N C E F O R M U L A
The distance between the points(x1, y1)and(x2, y2)is d =
(x2−x1)2+(y2−y1)2
(1)
We considered two points in the first quadrant to prove the distance formula. It turns out that the same formula is valid irrespective of where the two pointsP1andP2lie. Note also that since(x1−x2)2 =(x2−x1)2 and(y1 −y2)2 = (y2−y1)2, it makes no difference which point isP1and which isP2.
Some find formula (1) hard to grasp. In words it tells us that we can find the distance between two points in the plane as follows:
Take the difference between thex-coordinatesand square whatyou get.Do the same with they-coordinates. Add the resultsand then take the square root.
E X A M P L E 1 Find the distancedbetweenP1 =(−4,3)andP2 =(5,−1). (See Fig. 2.) Solution: Using (1) withx1= −4,y1=3 andx2 =5,y2= −1, we have
d=
(5−(−4))2+(−1−3)2 =
92+(−4)2 =
81+16=
97≈9.85
Circles
Let(a, b)be a point in the plane.The circle with radiusrand centreat(a, b)is the set of all points(x, y)whose distance from(a, b)is equal tor. Applying the distance formula to the typical point(x, y)on the circle shown in Fig. 3 gives
(x−a)2+(y−b)2 =r Squaring each side yields:
E Q U A T I O N O F A C I R C L E
The equation of a circle with centre at(a, b)and radiusris (x−a)2+(y−b)2=r2
(2)
A graph of (2) is shown in Fig. 3. Note that if we leta=b=0 andr =4, then (2) reduces tox2+y2 =16. This is the equation of a circle with centre at(0,0)and radius 4, as shown in Fig. 5.4.2.
r
(x, y)
(a, b) y
x
3 3
4
2 1
⫺1
⫺2
⫺2
⫺3
⫺4
⫺5
⫺6 ⫺1 1
y
x
Figure 3 Circle with centre at(a, b)and radiusr
Figure 4 Circle with centre at(−4,1) and radius 3
E X A M P L E 2 Find the equation of the circle with centre at(−4,1)and radius 3.
Solution: Herea = −4,b=1, andr=3. (See Fig. 4.) So according to (2), the equation for the circle is
(x+4)2+(y−1)2=9 (∗)
Expanding the squares to obtainx2+8x+16+y2−2y+1=9, and then collecting terms, we have
x2+y2+8x−2y+8=0 (∗∗)
The equation of the circle given in(∗∗)has the disadvantage that we cannot immediately read off its centre and radius. If we are given equation(∗∗), however, we can use the method of “completing the squares” to deduce(∗)from(∗∗). See Problem 5.
Ellipses and Hyperbolas
A very important type of curve in physics and astronomy is the ellipse. (After all, the planets, including the Earth, move around the Sun in orbits that are approximately elliptical.) Occasionally, ellipses also appear in economics and statistics. The simplest type of ellipse has the equation
(x−x0)2
a2 +(y−y0)2
b2 =1 (ellipse) (3)
This ellipse has centre at(x0, y0)and its graph is shown in Fig. 5. Note that whena =b, the ellipse degenerates into a circle.
b a
(x, y) y0
x0 y
x
a b y0
x0 y
x
b y0 a
x0 y
x
Figure 5 Ellipse Figure 6 Hyperbola Figure 7 Hyperbola
S E C T I O N 5 . 5 / D I S T A N C E I N T H E P L A N E . C I R C L E S 149
Figures 6 and 7 show the graphs of the twohyperbolas (x−x0)2
a2 −(y−y0)2
b2 = +1 and (x−x0)2
a2 −(y−y0)2
b2 = −1 (4) respectively. The asymptotes are the dashed lines in Figs. 6 and 7. They are the same pair in each figure. Their equations arey−y0= ±ab(x−x0).
We end this section by noting that the graph of the general quadratic equation
Ax2+Bxy+Cy2+Dx+Ey+F =0 (5)
whereA,B, andCare not all 0, will have one of the following shapes:
• If 4AC > B2, either an ellipse (possibly a circle), or a single point, or empty.
• If 4AC=B2, either a parabola, or one line or two parallel lines, or empty.
• If 4AC < B2, either a hyperbola, or two intersecting lines.
P R O B L E M S F O R S E C T I O N 5 . 5
1. Determine the distances between the following pairs of points:
(a) (1,3) and (2,4) (b) (−1,2) and (−3,3) (c) (3/2,−2)and(−5,1) (d) (x, y) and (2x, y+3) (e) (a, b) and (−a, b) (f) (a,3) and (2+a,5) 2. The distance between(2,4)and(5, y)is√
13. Findy. (Explain geometrically why there must be two values ofy.)
3. Find the distances between each pair of points:
(a) (3.998,2.114) and (1.130,−2.416) (b) (π,2π ) and (−π,1) 4. Find the equations of the circles with:
(a) Centre at(2,3)and radius 4. (b) Centre at(2,5)and one point at(−1,3).
5. To show that the graph ofx2+y2−10x+14y+58=0 is a circle, we can argue like this: First rearrange the equation to read(x2−10x)+(y2+14y)= −58. Completing the two squares gives:(x2−10x+52)+(y2+14y+72)= −58+52+72=16. Thus the equation becomes
(x−5)2+(y+7)2=16 whose graph is a circle with centre(5,−7)and radius√
16=4. Use this method to find the centre and the radius of the two circles with equations:
(a) x2+y2+10x−6y+30=0 (b) 3x2+3y2+18x−24y= −39
6. Prove that if the distance from a point(x, y)to the point(−2,0)is twice the distance from(x, y) to(4,0), then(x, y)must lie on the circle with centre(6,0)and radius 4.
7. In Example 4.7.7 we considered the functiony=(ax+b)/(cx+d), and we claimed that for c=0 the graph was a hyperbola. See how this accords with the classification below (5).
HARDER PROBLEM
⊂SM⊃8. Show that the graph of
x2+y2+Ax+By+C=0 (A,B, andCconstants) (∗) is a circle ifA2+B2 > 4C. Find its centre and radius. (See Problem 5.) What happens if A2+B2≤4C?