2 Annuli between circles
Equations in $x$ and $y$ , such as (1) i.e. ${x}^{2}+{y}^{2}={R}^{2}$ and (2) i.e. ${\left(x{x}_{0}\right)}^{2}+{\left(y{y}_{0}\right)}^{2}={R}^{2}$ for circles, define curves in the $Oxy$ plane. However, inequalities are necessary to define regions . For example, the inequality
$\phantom{\rule{2em}{0ex}}{x}^{2}+{y}^{2}<1$
is satisfied by all points inside the unit circle  for example $\left(0,0\right),\phantom{\rule{1em}{0ex}}\left(0,\frac{1}{2}\right),\phantom{\rule{1em}{0ex}}\left(\frac{1}{4},0\right),\phantom{\rule{1em}{0ex}}\left(\frac{1}{2},\frac{1}{2}\right)$ .
Similarly $\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{x}^{2}+{y}^{2}>1$ is satisfied by all points outside that circle such as $\left(1,1\right)$ .
Figure 31
Example 16
Sketch the regions in the $Oxy$ plane defined by
 ${\left(x1\right)}^{2}+{y}^{2}<1$
 ${\left(x1\right)}^{2}+{y}^{2}>1$
Solution
The equality $\phantom{\rule{1em}{0ex}}{\left(x1\right)}^{2}+{y}^{2}=1$ is satisfied by any point on the circumference of the circle centre (1,0) radius 1. Then, remembering that ${\left(x1\right)}^{2}+{y}^{2}$ is the square of the distance between any point $\left(x,y\right)$ and (1,0), it follows that
 ${\left(x1\right)}^{2}+{y}^{2}<1$ is satisfied by any point inside this circle (region (A) in the diagram.)

${\left(x1\right)}^{2}+{y}^{2}>1$
defines the region exterior to the circle since this inequality is satisfied by every point outside. (Region (B) on the diagram.)
The region between two circles with the same centre (i.e. concentric circles) is called an annulus or annular region . An annulus is defined by two inequalities. For example the inequality
$\phantom{\rule{2em}{0ex}}{x}^{2}+{y}^{2}>1$ (7)
defines, as we saw, the region outside the unit circle.
The inequality
$\phantom{\rule{2em}{0ex}}{x}^{2}+{y}^{2}<4$ (8)
defines the region inside the circle centre origin radius 2.
Hence points $\left(x,y\right)$ which satisfy both the inequalities (7) and (8) lie in the annulus between the two circles. The inequalities (7) and (8) are combined by writing
$\phantom{\rule{2em}{0ex}}1<{x}^{2}+{y}^{2}<4$
Figure 32
Task!
Sketch the annulus defined by the inequalities
The quantity ${\left(x1\right)}^{2}+{y}^{2}$ is the square of the distance of a point $\left(x,y\right)$ from the point (1,0). Hence, as we saw earlier, the lefthand inequality
$\phantom{\rule{2em}{0ex}}1<{\left(x1\right)}^{2}+{y}^{2}$ which is the same as $\phantom{\rule{2em}{0ex}}{\left(x1\right)}^{2}+{y}^{2}>1$
is the region exterior to the circle ${C}_{1}$ centre $\left(1,0\right)$ radius 1.
Similarly the righthand inequality
$\phantom{\rule{2em}{0ex}}{\left(x1\right)}^{2}+{y}^{2}<9$
defines the interior of the circle ${C}_{2}$ centre $\left(1,0\right)$ radius 3. Hence the double inequality holds for any point in the annulus between ${C}_{1}$ and ${C}_{2}$ .
Exercises

Write down the radius and the coordinates of the centre of the circle for each of the following equations
 ${x}^{2}+{y}^{2}=16$
 ${\left(x4\right)}^{2}+{\left(y3\right)}^{2}=12$
 ${\left(x+3\right)}^{2}+{\left(y1\right)}^{2}=25$
 ${x}^{2}+{\left(y+1\right)}^{2}4=0$
 ${\left(x+6\right)}^{2}+{y}^{2}36=0$

Obtain in each case the equation of the given circle
 centre $C\phantom{\rule{1em}{0ex}}\left(0,0\right)$ radius 7
 centre $C\phantom{\rule{1em}{0ex}}\left(0,2\right)$ radius 2
 centre $C\phantom{\rule{1em}{0ex}}\left(4,4\right)$ radius 4
 centre $C\phantom{\rule{1em}{0ex}}\left(2,2\right)$ radius 4
 centre $C\phantom{\rule{1em}{0ex}}\left(6,0\right)$ radius 5

Obtain the radius and the coordinates of the centre for each of the following circles
 ${x}^{2}+{y}^{2}10x+12y=0$
 ${x}^{2}+{y}^{2}+2x4y=11$
 ${x}^{2}+{y}^{2}6x16=0$

Describe the regions defined by each of these inequalities
 ${x}^{2}+{y}^{2}>4$
 ${x}^{2}+{y}^{2}<16$
 the inequalities in (i) and (ii) together
 State an inequality that describes the points that lie outside the circle of radius 4 with centre $\left(4,2\right)$ .
 State an inequality that describes the points that lie inside the circle of radius $\sqrt{6}$ with centre $\left(2,1\right)$ .
 Obtain the equation of the circle which has centre $\left(3,4\right)$ and which passes through the point $\left(0,5\right)$ .

Show that if
$A\left({x}_{1},{y}_{1}\right)$
and
$B\left({x}_{2},{y}_{2}\right)$
are at opposite ends of a diameter of a circle then the equation of the circle is
$\phantom{\rule{1em}{0ex}}\left(x{x}_{1}\right)\left(x{x}_{2}\right)+\left(y{y}_{1}\right)\left(y{y}_{2}\right)=0.$
(Hint: if $P$ is any point on the circle obtain the slopes of the lines $AP$ and $BP$ and recall that the angle in a semicircle must be a rightangle.)
 State the equation of the unique circle which touches the $x$ axis at the point (2,0) and which passes through the point $\left(1,9\right)$ .

 radius 4 centre $\left(0,0\right)$
 radius $\sqrt{12}$ centre $\left(4,3\right)$
 radius 5 centre $\left(3,1\right)$
 radius 2 centre $\left(0,1\right)$
 radius 6 centre $\left(6,0\right)$

 ${x}^{2}+{y}^{2}=49$
 ${x}^{2}+{\left(y2\right)}^{2}=4$
 ${\left(x4\right)}^{2}+{\left(y+4\right)}^{2}=16$
 ${\left(x+2\right)}^{2}+{\left(y+2\right)}^{2}=16$
 ${\left(x+6\right)}^{2}+{y}^{2}=25$

 centre $\left(5,6\right)$ radius $\sqrt{61}$
 centre $\left(1,2\right)$ radius 4
 centre (3,0) radius 5

 the region outside the circumference of the circle centre the origin radius 2.
 the region inside the circle centre the origin radius 4 (often referred to as a circular disc)
 the annular ring between these two circles.
 ${\left(x+4\right)}^{2}+{\left(y2\right)}^{2}>16$
 ${\left(x+2\right)}^{2}+{\left(y+1\right)}^{2}<6$
 ${\left(x3\right)}^{2}+{\left(y4\right)}^{2}=10$
 $\left(x{x}_{1}\right)\left(x{x}_{2}\right)+\left(y{y}_{1}\right)\left(y{y}_{2}\right)=0.$
 ${\left(x2\right)}^{2}+{\left(y5\right)}^{2}=25$ (Note: since we are told the circle touches the $x$ axis at (2,0) the centre of the circle must be at the point $\left(2,{y}_{0}\right)$ where ${y}_{0}=R$ ).