4.1 Angles and circles

Example 1

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

Example 2

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

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

2. The angles \displaystyle \frac{3\pi}{7} and \displaystyle -\frac{11\pi}{7} indicate the same direction 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 same direction, but opposite directions 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 the radius multiplied by the angle measured in radians,

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

2. Determine the area of the circle segment.
   
   The circle 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, i.e.

   \displaystyle \frac{5}{36} \cdot 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 gives 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 of the circle in the form

because then we can directly read from this that the midpoint 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).

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 equal to

and moving over the 4 to to the right-hand side we get the equation

We can interpret this as follows: The centre is at \displaystyle (1,-2) and the radius is \displaystyle \sqrt{4}= 2.