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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.

$chorddiameter :a segment drawn from one point on the circle to another.:a chord that intersects the center of the circle. $

Notice that a radius is half of the diameter and **not** a chord.
Additionally, there are two noteworthy lines that intersect a circle.
$secanttangent :a line that intersects the circle at two points.:a line that intersects the circle at exactly one point. $

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

Show Solution

To begin, notice that $m$ and $AD$ 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 $AB,$ $AC$ and $AD$ are all chords. Since $AC$ passes through the circle's center, $P,$ it's called a diameter. The segment $AD$ is a part of the line $AD.$ A line that intersects a circle at two points is called a secant. Thus, $AD$ is a secant.

The segment $PD$ 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 $PD.$

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

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

$AB$ | chord |

$AC$ | diameter |

$AD$ | secant |

$PD$ | 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 $Q.$

Line $m$ is tangent to $⊙Q$ $⇔$ $m⊥QP $

The theorem will be proven in two parts as it is a biconditional statement.

- If line $m$ is tangent to $⊙Q,$ then $m⊥QP .$
- If $m⊥QP ,$ then line $m$ is tangent to $⊙Q.$

Each part will be proven by contradiction. For the first part, it will be assumed that line $m$ is tangent to the circle and that $m$ is **not** perpendicular to $QP .$
$mis tangent to⊙QandQP ⊥ m $
By the Perpendicular Postulate, there is another segment from $Q$ that is perpendicular to $m.$ Let this segment be $QT .$ The aim is to show that $QP $ *must* be this segment.

Since $QT $ is perpendicular to $m,$ then $△QTP$ is a right triangle. In this right triangle, $QP ,$ whose length is $r$ units, is the hypotenuse and therefore the longest side. Consequently, $QT$ is *less than* $QP.$
$QT<QP $
In the diagram, it can be seen that $QT $ contains a radius of $⊙Q.$ Because part of the segment lies outside the circle, the unknown portion can be assigned another variable, $b.$

Line $m$ is tangent to $⊙Q$ $⇒$ $m⊥QP $

For the second part, it will be assumed that $m$ is perpendicular to the radius $QP $ at $P,$ and that line $m$ is **not** tangent to $⊙Q.$ In this case, line $m$ intersects $⊙Q$ at a second point $R.$

Since $m$ is perpendicular to $QP ,$ then $△QPR$ is a right triangle and $QR $ is the hypotenuse. Therefore, $QP$ must be *less than* $QR.$
$QP<QR $
However, both $QP $ and $QR $ are radii of $⊙Q.$ This means that they must have the same length.
$QP=QR $
These two statements contradict each other. Therefore, they **cannot** be both true. The contradiction came from supposing that line $m$ was **not** tangent to $⊙Q.$ Consequently, $m$ is a tangent line to the circle. This completes the second part of the proof.

$m⊥QP $ $⇒$ line $m$ is tangent to $⊙Q$

Having proven both parts completes the proof for the theorem.

Line $m$ is tangent to $⊙Q$ $⇔$ $m⊥QP $

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

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

Because $QT ⊥m,$ a right angle is created at their intersection. As a result, $QP ,$ which measures $r$ units, is the hypotenuse — and the longest side — in the right triangle $△QTP.$ Therefore, the length of $QT $ is shorter than $QP .$ $QT<QP$ Notice that $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.
$QT<QP⇔r+b<r.$
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 $m$ must be false. Therfore, $QP$ is perpendicular to the tangent $m.$

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

Statement | Reason |

$QP⊥ m$ | Given |

Draw line segment $QT$ | Construction of triangle |

$QT⊥m$ | Right triangle |

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

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

$QT<QP$ | Hypotenuse longer than leg |

$b≮0$ | Length must be positive |

$QP⊥m$ | Proof by contradiction |

All circles are similar.

In the diagram above, circle $B∼$ 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 $B.$
$k=r_{B}r_{A_{′}} .$
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 $B,$ $A∼B.$ Therefore, all circles are similar.

This can be summarized in the following flowchart proof.

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