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Intersecting Circles



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.

Radius circle.svg

Lines and Segments that Intersect Circles

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.

chord:a segment drawn from one point on the circle to another.diameter:a chord that intersects the center of the circle.\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. secant:a line that intersects the circle at two points.tangent:a line that intersects the circle at exactly one point.\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 PP is the circle's midpoint.

Show Solution

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

The segment PD\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,m, touches the circle at the point DD and is perpendicular to the radius PD.\overline{PD}.

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

line/segment name
AB\overline{AB} chord
AC\overline{AC} diameter
AD\overleftrightarrow{AD} secant
PD\overline{PD} radius
mm tangent

Tangent to Circle

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, mm is tangent to the circle Q.Q. This can be proven using the Perpendicular Postulate.


Tangent to Circle Theorem

According to the Perpendicular Postulate, there is only one segment from QQ that is perpendicular to m.\overleftrightarrow{m}. It will be shown, using contradiction, that QP\overline{QP} must be this segment.
Suppose that QP\overline{QP} is not perpendicular to m.\overleftrightarrow{m}. It follows then that there exists another segment from QQ perpendicular to m.\overleftrightarrow{m}. Call this segment QT.\overline{QT}.

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

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

Statement Reason
QPmQP \cancel{\perp} m Given
Draw line segment QTQT Construction of triangle
QTmQT\bot m Right triangle
QP=rQP=r Express triangle's hypotenuse
QT=r+bQT=r+b Express leg of triangle
QT<QPQT<QP Hypotenuse longer than leg
b0b\nless0 Length must be positive
QPmQP \bot m Proof by contradiction

Similar Circles Theorem

All circles are similar.

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


Similar Circles Theorem

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

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

Similarity transformation

To determine the scale factor, k,k, of the dilation, divide the radius of AA' with the radius of B.B. k=rArB. k=\frac{r_{A'}}{r_B}. Tha radius of AA' is then multiplied with the scale factor, which maps AA' to B.B. Thus, because a similarity transformation exists that maps circle AA onto circle B,B, AB.A \sim B. Therefore, all circles are similar.
This can be summarized in the following flowchart proof.

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