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.
Notice that a radius is half of the diameter and not a chord. Additionally, there are two noteworthy lines that intersect a circle.
Name the lines and segments in the circle where is the circle's midpoint.
To begin, notice that and 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 and are all chords. Since passes through the circle's center, it's called a diameter. The segment is a part of the line A line that intersects a circle at two points is called a secant. Thus, is a secant.
The segment 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, touches the circle at the point and is perpendicular to the radius
Therefore, is a tangent to the circle. We have now identified all lines and segments intersecting the circle.
In the diagram above, is tangent to the circle
Line is tangent to
The theorem will be proven in two parts as it is a biconditional statement.
Each part will be proven by contradiction. For the first part, it will be assumed that line is tangent to the circle and that is not perpendicular to By the Perpendicular Postulate, there is another segment from that is perpendicular to Let this segment be The aim is to show that must be this segment.
Since is perpendicular to then is a right triangle. In this right triangle, whose length is units, is the hypotenuse and therefore the longest side. Consequently, is less than In the diagram, it can be seen that contains a radius of Because part of the segment lies outside the circle, the unknown portion can be assigned another variable,
Line is tangent to
For the second part, it will be assumed that is perpendicular to the radius at and that line is not tangent to In this case, line intersects at a second point
Since is perpendicular to then is a right triangle and is the hypotenuse. Therefore, must be less than However, both and are radii of This means that they must have the same length. These two statements contradict each other. Therefore, they cannot be both true. The contradiction came from supposing that line was not tangent to Consequently, is a tangent line to the circle. This completes the second part of the proof.
line is tangent to
Having proven both parts completes the proof for the theorem.
Line is tangent to
According to the Perpendicular Postulate, there is only one segment from that is perpendicular to It will be shown, using contradiction, that must be this segment.
Suppose that is not perpendicular to It follows then that there exists another segment from perpendicular to Call this segment
Because a right angle is created at their intersection. As a result, which measures units, is the hypotenuse — and the longest side — in the right triangle Therefore, the length of is shorter than Notice that contains a radius of circle Because part of the segment lies outside the circle, the unknown portion can be assigned another variable,
By the Segment Addition Postulate, The relationship between and can be written as an inequality.
By subtracting on both sides, the inequality states that Since a length cannot be less than the assumption that is not perpendicular to must be false. Therfore, is perpendicular to the tangent
This can be summarized in the following two-column proof.
|Draw line segment||Construction of triangle|
|Express triangle's hypotenuse|
|Express leg of triangle|
|Hypotenuse longer than leg|
|Length must be positive|
|Proof by contradiction|
Two figures are similar if there exists a similarity transformation that maps one onto the other. Consider circles and as shown.
Circle can be mapped onto circle through a composite transformation. First, it can be translated left and down so that the midpoints overlap. Then the size of circle can be enlarged through a dilation.
To determine the scale factor, of the dilation, divide the radius of with the radius of
Tha radius of is then multiplied with the scale factor, which maps to Thus, because a similarity transformation exists that maps circle onto circle Therefore, all circles are similar.
This can be summarized in the following flowchart proof.