4.1 Angles and circles
From Förberedande kurs i matematik 1
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*To convert between degrees, radians and revolutions. | *To convert between degrees, radians and revolutions. | ||
*To calculate the area and circumference of sectors of a circle. | *To calculate the area and circumference of sectors of a circle. | ||
- | *The | + | *The features of right-angled triangles. |
*To formulate and use the Pythagorean theorem. | *To formulate and use the Pythagorean theorem. | ||
*To calculate the distance between two points in the plane. | *To calculate the distance between two points in the plane. | ||
- | *To sketch circles by completing the square | + | *To sketch circles by completing the square. |
- | *The concepts of unit circle, tangent, radius, diameter, circumference, chord and arc. | + | *The concepts of the unit circle, tangent, radius, diameter, circumference, chord and arc. |
*To solve geometric problems that contain circles. | *To solve geometric problems that contain circles. | ||
}} | }} | ||
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[[Image:Gradskiva - 57°.gif||center]] | [[Image:Gradskiva - 57°.gif||center]] | ||
- | *'''Radians.''' Another way to measure an angle, is to use the length of the arc | + | *'''Radians.''' Another way to measure an angle, is to use the length of the arc described by the angle in relation to the radius. This unit is called radian. A revolution is <math>2\pi</math> radians, since the circumference of a circle is <math>2\pi r</math>, where <math>r</math> is the radius of the circle. |
[[Image:Gradskiva - Radianer.gif||center]] | [[Image:Gradskiva - Radianer.gif||center]] | ||
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= \frac{\pi}{6}\ \mbox{ radians }</math></li> | = \frac{\pi}{6}\ \mbox{ radians }</math></li> | ||
<li><math>\frac{\pi}{8}\ \mbox { radians } | <li><math>\frac{\pi}{8}\ \mbox { radians } | ||
- | = \frac{\pi}{8} \cdot (1 \; \mbox{ | + | = \frac{\pi}{8} \cdot (1 \; \mbox{radian}\,) |
= \frac{\pi}{8} \cdot \frac{180^\circ}{\pi} | = \frac{\pi}{8} \cdot \frac{180^\circ}{\pi} | ||
- | = 22{ | + | = 22\mbox{.}5^\circ</math></li> |
</ol> | </ol> | ||
</div> | </div> | ||
- | In some contexts, it may be useful to talk about negative angles and angles greater than 360°. This means that the same | + | In some contexts, it may be useful to talk about negative angles and angles greater than 360°. This means that the same point on the circle can be designated by different angles that differ from each other by an integral number of revolutions. |
- | <center>{{:4.1 - | + | <center>{{:4.1 - Figure - Angles 45°, -315° and 405°}}</center> |
<div class="exempel"> | <div class="exempel"> | ||
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<ol type="a"> | <ol type="a"> | ||
<li> The angles <math>-55^\circ</math> and <math>665^\circ | <li> The angles <math>-55^\circ</math> and <math>665^\circ | ||
- | </math> indicate the same | + | </math> indicate the same point on the circle because |
{{Displayed math||<math> | {{Displayed math||<math> | ||
-55^\circ + 2 \cdot 360^\circ = 665^\circ\,\mbox{.}</math>}}</li> | -55^\circ + 2 \cdot 360^\circ = 665^\circ\,\mbox{.}</math>}}</li> | ||
<li> The angles <math>\frac{3\pi}{7}</math> and <math> | <li> The angles <math>\frac{3\pi}{7}</math> and <math> | ||
- | -\frac{11\pi}{7}</math> indicate the same | + | -\frac{11\pi}{7}</math> indicate the same point on the circle because |
{{Displayed math||<math> | {{Displayed math||<math> | ||
\frac{3\pi}{7} - 2\pi = -\frac{11\pi}{7}\,\mbox{.}</math>}}</li> | \frac{3\pi}{7} - 2\pi = -\frac{11\pi}{7}\,\mbox{.}</math>}}</li> | ||
<li> The angles <math>36^\circ</math> and <math> | <li> The angles <math>36^\circ</math> and <math> | ||
- | 216^\circ</math> do not specify the | + | 216^\circ</math> do not specify the samepoint on the circle, but opposite points since |
{{Displayed math||<math> | {{Displayed math||<math> | ||
36^\circ + 180^\circ = 216^\circ\,\mbox{.}</math>}}</li> | 36^\circ + 180^\circ = 216^\circ\,\mbox{.}</math>}}</li> | ||
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== Formula for distance in the plane == | == Formula for distance in the plane == | ||
- | The theorem of Pythagoras is one of the most famous theorems in mathematics and says that in a right-angled triangle with the legs <math>a</math> and <math>b</math>, and | + | The theorem of Pythagoras is one of the most famous theorems in mathematics and says that in a right-angled triangle with the legs <math>a</math> and <math>b</math>, and hypotenuse <math>c</math> then <math>a^2 + b^2 = c^2</math>. |
<div class="regel"> | <div class="regel"> | ||
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|width="100%"|'''The Pythagorean theorem:''' | |width="100%"|'''The Pythagorean theorem:''' | ||
{{Displayed math||<math>c^2 = a^2 + b^2\,\mbox{.}</math>}} | {{Displayed math||<math>c^2 = a^2 + b^2\,\mbox{.}</math>}} | ||
- | |align="right"|{{:4.1 - | + | |align="right"|{{:4.1 - Figure - The Pythagorean theorem}} |
|} | |} | ||
</div> | </div> | ||
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{| width="100%" | {| width="100%" | ||
- | |width="100%"| | + | |width="100%"| Consider the triangle on the right. Then |
{{Displayed math||<math>c^2= 3^2 + 4^2 = 9 +16 = 25</math>}} | {{Displayed math||<math>c^2= 3^2 + 4^2 = 9 +16 = 25</math>}} | ||
- | and therefore hypotenuse <math>c</math> is equal to | + | and therefore the hypotenuse <math>c</math> is equal to |
{{Displayed math||<math>c=\sqrt{25} = 5\,\mbox{.}</math>}} | {{Displayed math||<math>c=\sqrt{25} = 5\,\mbox{.}</math>}} | ||
- | |align="right"|{{:4.1 - | + | |align="right"|{{:4.1 - Figure - A right-angled triangle with sides 3, 4 and 5}} |
|} | |} | ||
</div> | </div> | ||
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- | The distance <math>r</math> is called the | + | The distance <math>r</math> is called the circle's radius and the point <math>(a,b)</math> is its centre. The figure below shows the other important concepts. |
{| align="center" | {| align="center" | ||
- | |align="center" valign="bottom"|{{:4.1 - | + | |align="center" valign="bottom"|{{:4.1 - Figure - Diameter}} |
|width="15px"| | |width="15px"| | ||
- | |align="center" valign="bottom"|{{:4.1 - | + | |align="center" valign="bottom"|{{:4.1 - Figure - Tangent}} |
|width="15px"| | |width="15px"| | ||
- | |align="center" valign="bottom"|{{:4.1 - | + | |align="center" valign="bottom"|{{:4.1 - Figure - The chord of a circle}} |
|width="15px"| | |width="15px"| | ||
- | |align="center" valign="bottom"|{{:4.1 - | + | |align="center" valign="bottom"|{{:4.1 - Figure - Secant}} |
|- | |- | ||
|align="center"|Diameter | |align="center"|Diameter | ||
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|height="15px"| | |height="15px"| | ||
|- | |- | ||
- | |align="center"|{{:4.1 - | + | |align="center"|{{:4.1 - Figure - Arc of a circle}} |
|| | || | ||
- | |align="center"|{{:4.1 - | + | |align="center"|{{:4.1 - Figure - Circumference}} |
|| | || | ||
- | |align="center"|{{:4.1 - | + | |align="center"|{{:4.1 - Figure - Sector of a circle}} |
|| | || | ||
- | |align="center"|{{:4.1 - | + | |align="center"|{{:4.1 - Figure - Segment of a circle}} |
|- | |- | ||
|align="center"| Arc of a circle | |align="center"| Arc of a circle | ||
|| | || | ||
- | |align="center"| | + | |align="center"| Circumference |
|| | || | ||
- | |align="center"| | + | |align="center"| Sector of a circle |
|| | || | ||
- | |align="center"| | + | |align="center"|Segment of a circle |
|} | |} | ||
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{| width="100%" | {| width="100%" | ||
- | ||A sector of a circle is | + | ||A sector of a circle is shown in the figure on the right. |
<ol type="a"> | <ol type="a"> | ||
<li> Determine its arc length . | <li> Determine its arc length . | ||
<br> | <br> | ||
<br> | <br> | ||
- | The central angle <math>50^\circ</math> | + | The central angle is <math>50^\circ</math>. In radians this is |
{{Displayed math||<math> | {{Displayed math||<math> | ||
50^\circ = 50 \cdot 1^\circ | 50^\circ = 50 \cdot 1^\circ | ||
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</ol> | </ol> | ||
|align="right" valign="top"| | |align="right" valign="top"| | ||
- | {{:4.1 - | + | {{:4.1 - Figure - 50° sector of a circle}} |
|} | |} | ||
<ol style="list-style-type:none; padding-top:0; margin-top:0;"> | <ol style="list-style-type:none; padding-top:0; margin-top:0;"> | ||
- | <li>The way radians have been defined means that the arc length is the radius multiplied by the angle measured in radians, | + | <li>The way radians have been defined means that the arc length is the radius multiplied by the angle measured in radians, so that the arc length is |
{{Displayed math||<math> | {{Displayed math||<math> | ||
3 \cdot \frac{5\pi}{18}\ \mbox{units } | 3 \cdot \frac{5\pi}{18}\ \mbox{units } | ||
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<br> | <br> | ||
<br> | <br> | ||
- | The | + | The segment's share of the entire circle is |
{{Displayed math||<math> | {{Displayed math||<math> | ||
\frac{50^\circ}{360^\circ} = \frac{5}{36}</math>}} | \frac{50^\circ}{360^\circ} = \frac{5}{36}</math>}} | ||
- | and this means that its area is <math>\frac{5}{36}</math> parts of the circle area, which is <math>\pi r^2 = \pi 3^2 = 9\pi</math> | + | and this means that its area is <math>\frac{5}{36}</math> parts of the circle area, which is <math>\pi r^2 = \pi 3^2 = 9\pi</math>. Hence the area is |
{{Displayed math||<math> | {{Displayed math||<math> | ||
\frac{5}{36} \cdot 9\pi\ \mbox{ units }= \frac{5\pi}{4}\ \mbox{ units. }</math>}}</li> | \frac{5}{36} \cdot 9\pi\ \mbox{ units }= \frac{5\pi}{4}\ \mbox{ units. }</math>}}</li> | ||
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</div> | </div> | ||
- | A point <math>(x,y)</math> lies on the circle that has its centre at <math>(a,b)</math> and radius <math>r</math>, if its distance from the centre is equal to <math>r</math>. This condition can be formulated with the distance formula | + | A point <math>(x,y)</math> lies on the circle that has its centre at <math>(a,b)</math> and radius <math>r</math>, if its distance from the centre is equal to <math>r</math>. This condition can be formulated with the distance formula. |
<div class="regel"> | <div class="regel"> | ||
{| width="100%" | {| width="100%" | ||
||'''Circle equation: ''' | ||'''Circle equation: ''' | ||
{{Displayed math||<math>(x – a)^2 + (y – b)^2 = r^2\,\mbox{.}</math>}} | {{Displayed math||<math>(x – a)^2 + (y – b)^2 = r^2\,\mbox{.}</math>}} | ||
- | |align="right"|{{:4.1 - | + | |align="right"|{{:4.1 - Figure - Equation of a circle}} |
|} | |} | ||
</div> | </div> | ||
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|width="100%"| | |width="100%"| | ||
<ol type="a"> | <ol type="a"> | ||
- | <li><math>(x-1)^2 + (y-2)^2 = 9\quad</math> is the equation for a circle with | + | <li><math>(x-1)^2 + (y-2)^2 = 9\quad</math> is the equation for a circle with centre <math>(1,2)</math> and radius <math>\sqrt{9} = 3</math>.</li> |
</ol> | </ol> | ||
- | |align="right"|{{:4.1 - | + | |align="right"|{{:4.1 - Figure - The equation (x - 1)² + (y - 2)² = 9}} |
|- | |- | ||
|width="100%"| | |width="100%"| | ||
<ol type="a" start=2> | <ol type="a" start=2> | ||
- | <li><math>x^2 + (y-1)^2 = 1\quad</math> can be written as <math>(x-0)^2 + (y-1)^2 = 1</math> and is the equation of a circle with | + | <li><math>x^2 + (y-1)^2 = 1\quad</math> can be written as <math>(x-0)^2 + (y-1)^2 = 1</math> and is the equation of a circle with centre <math>(0,1)</math> and radius <math>\sqrt{1} = 1</math>.</li> |
</ol> | </ol> | ||
- | |align="right"|{{:4.1 - | + | |align="right"|{{:4.1 - Figure - The equation x² + (y - 1)² = 1}} |
|- | |- | ||
|width="100%"| | |width="100%"| | ||
<ol type="a" start=3> | <ol type="a" start=3> | ||
- | <li><math>(x+1)^2 + (y-3)^2 = 5\quad</math> can be written as <math>(x-(-1))^2 + (y-3)^2 = 5</math> and is the equation of a circle with | + | <li><math>(x+1)^2 + (y-3)^2 = 5\quad</math> can be written as <math>(x-(-1))^2 + (y-3)^2 = 5</math> and is the equation of a circle with centre <math>(-1,3)</math> and radius <math>\sqrt{5} \approx 2\textrm{.}236</math>.</li> |
</ol> | </ol> | ||
- | |align="right"|{{:4.1 - | + | |align="right"|{{:4.1 - Figure - The equation (x + 1)² + (y - 3)² = 5}} |
|} | |} | ||
</div> | </div> | ||
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\end{align*}</math>}} | \end{align*}</math>}} | ||
Since the point satisfies the circle equation it lies on the circle. | Since the point satisfies the circle equation it lies on the circle. | ||
- | <center>{{:4.1 - | + | <center>{{:4.1 - Figure - The equation (x - 4)² + y² = 13}}</center></li> |
<li> Determine the equation for the circle that has its centre at <math>(3,4)</math> and goes through the point <math>(1,0)</math>. | <li> Determine the equation for the circle that has its centre at <math>(3,4)</math> and goes through the point <math>(1,0)</math>. | ||
<br> | <br> | ||
<br> | <br> | ||
- | Since the point <math>(1,0)</math> lies on the circle, the radius of the circle must be equal to the distance of the point from <math>(1,0)</math> to the centre <math>(3,4)</math>. The distance formula | + | Since the point <math>(1,0)</math> lies on the circle, the radius of the circle must be equal to the distance of the point from <math>(1,0)</math> to the centre <math>(3,4)</math>. The distance formula allows us to calculate that this distance is |
{{Displayed math||<math> | {{Displayed math||<math> | ||
c = \sqrt{(3-1)^2 + (4-0)^2} = \sqrt{4 +16} = \sqrt{20} \, \mbox{.}</math>}} | c = \sqrt{(3-1)^2 + (4-0)^2} = \sqrt{4 +16} = \sqrt{20} \, \mbox{.}</math>}} | ||
The circle equation is therefore | The circle equation is therefore | ||
{{Displayed math||<math>(x-3)^2 + (y-4)^2 = 20 \; \mbox{.}</math>}} | {{Displayed math||<math>(x-3)^2 + (y-4)^2 = 20 \; \mbox{.}</math>}} | ||
- | <center>{{:4.1 - | + | <center>{{:4.1 - Figure - The equation (x - 3)² + (y - 4)² = 20}}</center></li> |
</ol> | </ol> | ||
</div> | </div> | ||
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- | Let us try to write the equation | + | Let us try to write the equation in the form |
{{Displayed math||<math>(x – a)^2 + (y – b)^2 = r^2</math>}} | {{Displayed math||<math>(x – a)^2 + (y – b)^2 = r^2</math>}} | ||
- | because then we can directly read from this that the | + | because then we can directly read from this that the centre is <math>(a,b)</math> and the radius is <math>r</math>. |
Start by completing the square for the terms containing <math>x</math> on the left-hand side | Start by completing the square for the terms containing <math>x</math> on the left-hand side | ||
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\underline{x^2-2x\vphantom{(}} + y^2+4y + 1 | \underline{x^2-2x\vphantom{(}} + y^2+4y + 1 | ||
= \underline{(x-1)^2-1^2} + y^2+4y + 1</math>}} | = \underline{(x-1)^2-1^2} + y^2+4y + 1</math>}} | ||
- | + | The underlined terms shows the terms involved. | |
- | + | Now complete the square for the terms containing <math>y</math> | |
{{Displayed math||<math> | {{Displayed math||<math> | ||
(x-1)^2-1^2 + \underline{y^2+4y} + 1 | (x-1)^2-1^2 + \underline{y^2+4y} + 1 | ||
= (x-1)^2-1^2 + \underline{(y+2)^2-2^2} + 1\,\mbox{.}</math>}} | = (x-1)^2-1^2 + \underline{(y+2)^2-2^2} + 1\,\mbox{.}</math>}} | ||
- | The left-hand side is equal to | + | The left-hand side is therefore equal to |
- | {{Displayed math||<math> (x-1)^2 + (y+2)^2-4 </math>}} | + | {{Displayed math||<math> (x-1)^2 + (y+2)^2-4 </math>.}} |
- | + | Rearranging we see that the original equation is equivalent to | |
{{Displayed math||<math> (x-1)^2 + (y+2)^2 = 4 \, \mbox{.}</math>}} | {{Displayed math||<math> (x-1)^2 + (y+2)^2 = 4 \, \mbox{.}</math>}} | ||
- | + | From this we can easily see that the centre is <math>(1,-2)</math> and the radius is <math>\sqrt{4}= 2</math>. | |
- | <center>{{:4.1 - | + | <center>{{:4.1 - Figure - The equation x² + y² - 2x + 4y + 1 = 0}}</center> |
</div> | </div> | ||
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- | '''Keep in mind that | + | '''Keep in mind that...''' |
'''Reviews''' | '''Reviews''' | ||
- | For those of you who want to deepen your | + | For those of you who want to deepen your understanding or need more detailed explanations consider the following references: |
- | [http://sv.wikipedia.org/wiki/Pythagoras_sats Learn more about Pythagoras theorem | + | [http://sv.wikipedia.org/wiki/Pythagoras_sats Learn more about Pythagoras theorem from Wikipedia ] |
- | [http://mathworld.wolfram.com/Circle.html Read more | + | [http://mathworld.wolfram.com/Circle.html Read more about the circle on the Mathworld website] |
Current revision
Theory | Exercises |
Contents:
- Various angle measures (degrees, radians and revolutions)
- The Pythagorean theorem
- Formula for distance in the plane
- Equation of a circle
Learning outcomes:
After this section, you will have learned :
- To convert between degrees, radians and revolutions.
- To calculate the area and circumference of sectors of a circle.
- The features of right-angled triangles.
- To formulate and use the Pythagorean theorem.
- To calculate the distance between two points in the plane.
- To sketch circles by completing the square.
- The concepts of the unit circle, tangent, radius, diameter, circumference, chord and arc.
- To solve geometric problems that contain circles.
Angle measures
There are several different units for measuring angles, which are used in different contexts. The two most common within mathematics are degrees and radians.
- Degrees. If a complete revolution is divided into 360 parts, then each part is called 1 degree. Degrees are designated by \displaystyle {}^\circ.
- Radians. Another way to measure an angle, is to use the length of the arc described by the angle in relation to the radius. This unit is called radian. A revolution is \displaystyle 2\pi radians, since the circumference of a circle is \displaystyle 2\pi r, where \displaystyle r is the radius of the circle.
A complete revolution is \displaystyle 360^\circ or \displaystyle 2\pi radians which means
\displaystyle \begin{align*}
&1^\circ = \frac{1}{360} \cdot 2\pi\ \mbox{ radians } = \frac{\pi}{180}\ \mbox{ radians,}\\ &1\ \mbox{ radian } = \frac{1}{2\pi} \cdot 360^\circ = \frac{180^\circ}{\pi}\,\mbox{.} \end{align*} |
These conversion relations can be used to convert between degrees and radians.
Example 1
- \displaystyle 30^\circ = 30 \cdot 1^\circ = 30 \cdot \frac{\pi}{180}\ \mbox{ radians } = \frac{\pi}{6}\ \mbox{ radians }
- \displaystyle \frac{\pi}{8}\ \mbox { radians } = \frac{\pi}{8} \cdot (1 \; \mbox{radian}\,) = \frac{\pi}{8} \cdot \frac{180^\circ}{\pi} = 22\mbox{.}5^\circ
In some contexts, it may be useful to talk about negative angles and angles greater than 360°. This means that the same point on the circle can be designated by different angles that differ from each other by an integral number of revolutions.
Example 2
- The angles \displaystyle -55^\circ and \displaystyle 665^\circ
indicate the same point on the circle because
\displaystyle -55^\circ + 2 \cdot 360^\circ = 665^\circ\,\mbox{.}
- 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{.}
- 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{.}
Formula for distance in the plane
The theorem of Pythagoras is one of the most famous theorems in mathematics and says that in a right-angled triangle with the legs \displaystyle a and \displaystyle b, and hypotenuse \displaystyle c then \displaystyle a^2 + b^2 = c^2.
The Pythagorean theorem:
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Example 3
Consider the triangle on the right. Then
and therefore the hypotenuse \displaystyle c is equal to
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The Pythagorean theorem can be used to calculate the distance between two points in a coordinate system.
Formula for distance:
The distance \displaystyle d between two points with coordinates \displaystyle (x,y) and \displaystyle (a,b) is
\displaystyle d = \sqrt{(x – a)^2 + (y – b)^2}\,\mbox{.} |
The line joining the points is the hypotenuse of a triangle whose legs are parallel to the coordinate axes.
The legs of the triangle have lengths equal to the difference in the x- and y-directions of the points, that is \displaystyle |x-a| and \displaystyle |y-b|. The Pythagorean theorem then gives the formula for the distance.
Example 4
- 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{.}
- 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{.}
Circles
A circle consists of all the points that are at a given fixed distance \displaystyle r from a point \displaystyle (a,b).
The distance \displaystyle r is called the circle's radius and the point \displaystyle (a,b) is its centre. The figure below shows the other important concepts.
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Diameter | Tangent | Chord | Secant | |||
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Arc of a circle | Circumference | Sector of a circle | Segment of a circle |
Example 5
A sector of a circle is shown in the figure on the right.
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- The way radians have been defined means that the arc length is the radius multiplied by the angle measured in radians, so that the arc length is
\displaystyle 3 \cdot \frac{5\pi}{18}\ \mbox{units } = \frac{5\pi}{6}\ \mbox{ units . }
- 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} \cdot 9\pi\ \mbox{ units }= \frac{5\pi}{4}\ \mbox{ units. }
A point \displaystyle (x,y) lies on the circle that has its centre at \displaystyle (a,b) and radius \displaystyle r, if its distance from the centre is equal to \displaystyle r. This condition can be formulated with the distance formula.
Circle equation:
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Example 6
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Example 7
- 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.
- 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
\displaystyle (x – a)^2 + (y – b)^2 = r^2 |
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
\displaystyle
\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
\displaystyle
(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
\displaystyle (x-1)^2 + (y+2)^2-4 . |
Rearranging we see that the original equation is equivalent to
\displaystyle (x-1)^2 + (y+2)^2 = 4 \, \mbox{.} |
From this we can easily see that the centre is \displaystyle (1,-2) and the radius is \displaystyle \sqrt{4}= 2.
Study advice
The basic and final tests
After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.
Keep in mind that...
Reviews
For those of you who want to deepen your understanding or need more detailed explanations consider the following references:
Learn more about Pythagoras theorem from Wikipedia
Read more about the circle on the Mathworld website
Useful web sites
Interactive experiments: the sine and cosine on the unit circle (Flash)