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{{ printedBook.courseTrack.name }} {{ printedBook.name }} A circle is the set of all points that are equidistant from a given point. The given point is usually called the *center* of the circle. The distance between the center and any point on the circle is called a radius.

In a circle, segments can be drawn from the circle and the center. Depending how the segments are drawn, they have different names and properties.

$\begin{aligned} \textbf{chord} &: \text{a segment drawn from one point on the circle to another.} \\ \textbf{diameter} &: \text{a chord that intersects the center of the circle.} \\ \end{aligned}$

Notice that a radius is half of the diameter and **not** a chord.
Additionally, there are two noteworthy lines that intersect a circle.
$\begin{aligned}
\textbf{secant} &: \text{a line that intersects the circle at two points.}\\
\textbf{tangent} &: \text{a line that intersects the circle at exactly one point.}
\end{aligned}$

Name the lines and segments in the circle where $P$ is the circle's midpoint.

Show Solution

To begin, notice that $\overleftrightarrow{m}$ and $\overleftrightarrow{AD}$ are the only lines in the figure. Since the others are segments, it is possible that they are chords. Recall that a chord is a segment between two points on the circle. Thus, we can identify three chords in this circle.

Therefore, the segments $\overline{AB},$ $\overline{AC}$ and $\overline{AD}$ are all chords. Since $\overline{AC}$ passes through the circle's center, $P,$ it's called a diameter. The segment $\overline{AD}$ is a part of the line $\overleftrightarrow{AD}.$ A line that intersects a circle at two points is called a secant. Thus, $\overleftrightarrow{AD}$ is a secant.

The segment $\overline{PD}$ has endpoints at the circle and its center. Such segment is called a radius and is half the length of a diameter. The last line, $m,$ touches the circle at the point $D$ and is perpendicular to the radius $\overline{PD}.$

Therefore, $m$ is a tangent to the circle. We have now identified all lines and segments intersecting the circle.

line/segment | name |
---|---|

$\overline{AB}$ | chord |

$\overline{AC}$ | diameter |

$\overleftrightarrow{AD}$ | secant |

$\overline{PD}$ | radius |

$m$ | tangent |

A line is tangent to a circle if and only if the line is perpendicular to a radius of the circle.

In the diagram above, $m$ is tangent to the circle $Q.$ This can be proven using the Perpendicular Postulate.

According to the Perpendicular Postulate, there is only one segment from $Q$ that is perpendicular to $\overleftrightarrow{m}.$ It will be shown, using contradiction, that $\overline{QP}$ must be this segment.

Suppose that $\overline{QP}$ is **not** perpendicular to $\overleftrightarrow{m}.$ It follows then that there exists another segment from $Q$ perpendicular to $\overleftrightarrow{m}.$ Call this segment $\overline{QT}.$

Because $\overline{QT} \perp \overleftrightarrow{m},$ a right angle is created at their intersection. As a result, $\overline{QP},$ which measures $r$ units, is the hypotenuse — and the longest side — in the right triangle $\triangle QTP.$ Therefore, the length of $\overline{QT}$ is shorter than $\overline{QP}.$ $QT < QP$ Notice that $\overline{QT}$ contains a radius of circle $Q.$ Because part of the segment lies outside the circle, the unknown portion can be assigned another variable, $b.$

By the Segment Addition Postulate, $QT=r+b.$ The relationship between $QT$ and $QP$ can be written as an inequality.
$QT<QP \Leftrightarrow r+b<r.$
By subtracting $r$ on both sides, the inequality states that $b<0.$ Since a length cannot be less than $0,$ the assumption that $QP$ is not perpendicular to $\overleftrightarrow{m}$ must be false. Therfore, $QP$ is perpendicular to the tangent $m.$

This can be summarized in the following two-column proof.

Statement | Reason |

$QP \cancel{\perp} m$ | Given |

Draw line segment $QT$ | Construction of triangle |

$QT\bot m$ | Right triangle |

$QP=r$ | Express triangle's hypotenuse |

$QT=r+b$ | Express leg of triangle |

$QT<QP$ | Hypotenuse longer than leg |

$b\nless0$ | Length must be positive |

$QP \bot m$ | Proof by contradiction |

All circles are similar.

In the diagram above, circle $B \sim$ circle $A.$ This can be proven using similarity transformations.

Two figures are similar if there exists a similarity transformation that maps one onto the other. Consider circles $A$ and $B$ as shown.

Circle $A$ can be mapped onto circle $B$ through a composite transformation. First, it can be translated left and down so that the midpoints overlap. Then the size of circle $A$ can be enlarged through a dilation.

Similarity transformation

To determine the scale factor, $k,$ of the dilation, divide the radius of $A'$ with the radius of $B.$
$k=\frac{r_{A'}}{r_B}.$
Tha radius of $A'$ is then multiplied with the scale factor, which maps $A'$ to $B.$ Thus, because a similarity transformation exists that maps circle $A$ onto circle $B,$ $A \sim B.$ Therefore, all circles are similar.

This can be summarized in the following flowchart proof.

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