4.1 Angles and circles

Example 1

1. \displaystyle 30^\circ = 30 \times 1^\circ = 30 \times \frac{\pi}{180}\ \mbox{ radians } = \frac{\pi}{6}\ \mbox{ radians }
2. \displaystyle \frac{\pi}{8}\ \mbox { radians } = \frac{\pi}{8} \times (1 \; \mbox{radian}\,) = \frac{\pi}{8} \times \frac{180^\circ}{\pi} = 22\mbox{.}5^\circ

Example 2

1. The angles \displaystyle -55^\circ and \displaystyle 665^\circ indicate the same point on the circle because

   \displaystyle -55^\circ + 2 \times 360^\circ = 665^\circ\,\mbox{.}

2. The angles \displaystyle \frac{3\pi}{7} and \displaystyle -\frac{11\pi}{7} indicate the same point on the circle because

   \displaystyle \frac{3\pi}{7} - 2\pi = -\frac{11\pi}{7}\,\mbox{.}

3. The angles \displaystyle 36^\circ and \displaystyle 216^\circ do not specify the samepoint on the circle, but opposite points since

   \displaystyle 36^\circ + 180^\circ = 216^\circ\,\mbox{.}

Example 3

Example 4

1. The distance between \displaystyle (1,2) and \displaystyle (3,1) is

   \displaystyle d = \sqrt{ (1-3)^2 + (2-1)^2} = \sqrt{(-2)^2 + 1^2} = \sqrt{ 4+1} = \sqrt{5}\,\mbox{.}

2. The distance between \displaystyle (-1,0) and \displaystyle (-2,-5) is

   \displaystyle d = \sqrt{ (-1-(-2))^2 + (0-(-5))^2} = \sqrt{1^2 + 5^2} = \sqrt{1+25} = \sqrt{26}\,\mbox{.}

Example 5

1. The way radians have been defined means that the arc length is simply the radius multiplied by the angle measured in radians, so that the arc length is

   \displaystyle 3 \times \frac{5\pi}{18}\ \mbox{units } = \frac{5\pi}{6}\ \mbox{ units . }

2. Determine the area of the circle segment.
   
   The segment's share of the entire circle is

   \displaystyle \frac{50^\circ}{360^\circ} = \frac{5}{36}

   and this means that its area is \displaystyle \frac{5}{36} parts of the circle area, which is \displaystyle \pi r^2 = \pi 3^2 = 9\pi. Hence the area is

   \displaystyle \frac{5}{36} \times 9\pi\ \mbox{ units }= \frac{5\pi}{4}\ \mbox{ units. }

Example 6

Example 7

1. Does the point \displaystyle (1,2) lie on the circle \displaystyle (x-4)^2 +y^2=13?
   
   Inserting the coordinates of the point \displaystyle x=1 and \displaystyle y=2 in the circle equation, we have that

   \displaystyle \begin{align*} \mbox{LHS } &= (1-4)^2+2^2\\ &= (-3)^2+2^2 = 9+4 = 13 = \mbox{RHS}\,\mbox{.} \end{align*}

   Since the point satisfies the circle equation it lies on the circle.

   

2. Determine the equation for the circle that has its centre at \displaystyle (3,4) and goes through the point \displaystyle (1,0).
   
   Since the point \displaystyle (1,0) lies on the circle, the radius of the circle must be equal to the distance of the point from \displaystyle (1,0) to the centre \displaystyle (3,4). The distance formula allows us to calculate that this distance is

   \displaystyle c = \sqrt{(3-1)^2 + (4-0)^2} = \sqrt{4 +16} = \sqrt{20} \, \mbox{.}

   The circle equation is therefore

   \displaystyle (x-3)^2 + (y-4)^2 = 20 \; \mbox{.}

   

Example 8

Determine the centre and radius of the circle with equation \displaystyle \ x^2 + y^2 – 2x + 4y + 1 = 0.

Let us try to write the equation in the form

because then we can directly read from this that the centre is \displaystyle (a,b) and the radius is \displaystyle r.

Start by completing the square for the terms containing \displaystyle x on the left-hand side

 \underline{x^2-2x\vphantom{(}} + y^2+4y + 1
 = \underline{(x-1)^2-1^2} + y^2+4y + 1

The underlined terms shows the terms involved.

Now complete the square for the terms containing \displaystyle y

 (x-1)^2-1^2 + \underline{y^2+4y} + 1
 = (x-1)^2-1^2 + \underline{(y+2)^2-2^2} + 1\,\mbox{.}

The left-hand side is therefore equal to

Rearranging we see that the original equation is equivalent to

From this we can easily see that the centre is \displaystyle (1,-2) and the radius is \displaystyle \sqrt{4}= 2.