Given two points $(x_1,y_1)$ and $(x_2,y_2)$, recall that their horizontal distance from one another is $\Delta x=x_2-x_1$ and their vertical distance from one another is $\Delta y=y_2-y_1$. (Actually, the word "distance'' normally denotes "positive distance''. $\Delta x$ and $\Delta y$ are signed distances, but this is clear from context.) The actual (positive) distance from one point to the other is the length of the hypotenuse of a right triangle with legs $|\Delta x|$ and $|\Delta y|$, as shown in figure 1.2.1. The Pythagorean theorem then says that the distance between the two points is the square root of the sum of the squares of the horizontal and vertical sides: $$ \hbox{distance} =\sqrt{(\Delta x)^2+(\Delta y)^2}=\sqrt{(x_2-x_1)^2+ (y_2-y_1)^2}. $$ For example, the distance between points $A(2,1)$ and $B(3,3)$ is $\sqrt{(3-2)^2+(3-1)^2}=\sqrt{5}$.

Figure 1.2.1. Distance between two points, $\Delta x$ and $\Delta y$ positive.

As a special case of the distance formula, suppose we want to know the distance of a point $(x,y)$ to the origin. According to the distance formula, this is $\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}$.

A point $(x,y)$ is at a distance $r$ from the origin if and only if $\sqrt{x^2+y^2}=r$, or, if we square both sides: $x^2+y^2=r^2$. This is the equation of the circle of radius $r$ centered at the origin. The special case $r=1$ is called the unit circle; its equation is $x^2+y^2=1$.

Similarly, if $C(h,k)$ is any fixed point, then a point $(x,y)$ is at a distance $r$ from the point $C$ if and only if $\sqrt{(x-h)^2+(y-k)^2}=r$, i.e., if and only if $$ (x-h)^2+(y-k)^2=r^2. $$ This is the equation of the circle of radius $r$ centered at the point $(h,k)$. For example, the circle of radius 5 centered at the point $(0,-6)$ has equation $(x-0)^2+(y- -6)^2=25$, or $x^2+(y+6)^2=25$. If we expand this we get $x^2+y^2+12y+36=25$ or $x^2+y^2+12y+11=0$, but the original form is usually more useful.

Example 1.2.1 Graph the circle $x^2-2x+y^2+4y-11=0$. With a little thought we convert this to $(x-1)^2+(y+2)^2-16=0$ or $(x-1)^2+(y+2)^2=16$. Now we see that this is the circle with radius 4 and center $(1,-2)$, which is easy to graph. $\square$

Exercises 1.2

Ex 1.2.1 Find the equation of the circle of radius 3 centered at:

    a) $(0,0)$d) $(0,3)$
    b) $(5,6)$e) $(0,-3)$
    c) $(-5,-6)$f) $(3,0)$


Ex 1.2.2 For each pair of points $A(x_1,y_1)$ and $B(x_2,y_2)$ find (i) $\Delta x$ and $\Delta y$ in going from $A$ to $B$, (ii) the slope of the line joining $A$ and $B$, (iii) the equation of the line joining $A$ and $B$ in the form $y=mx+b$, (iv) the distance from $A$ to $B$, and (v) an equation of the circle with center at $A$ that goes through $B$.

    a) $A(2,0)$, $B(4,3)$d) $A(-2,3)$, $B(4,3)$
    b) $A(1,-1)$, $B(0,2)$e) $A(-3,-2)$, $B(0,0)$
    c) $A(0,0)$, $B(-2,-2)$f) $A(0.01,-0.01)$, $B(-0.01,0.05)$


Ex 1.2.3 Graph the circle $x^2+y^2+10y=0$.

Ex 1.2.4 Graph the circle $x^2-10x+y^2=24$.

Ex 1.2.5 Graph the circle $x^2-6x+y^2-8y=0$.

Ex 1.2.6 Find the standard equation of the circle passing through $(-2,1)$ and tangent to the line $3x-2y =6$ at the point $(4,3)$. Sketch. (Hint: The line through the center of the circle and the point of tangency is perpendicular to the tangent line.) (answer)