- The distance from the center of the circle to the straight line is greater than the radius. In this case, all points of the line lie outside the circle.
- The distance from the center of the circle to the straight line is less than the radius. In this case, the straight line has points lying inside the circle and since the straight line is infinite in both directions, it is intersected by the circle at 2 points.
- The distance from the center of the circle to the straight line is equal to the radius. Straight line is tangent.
A straight line that has only one point in common with a circle is called tangent to the circle.
The common point is called in this case point of contact.
The possibility of the existence of a tangent, and, moreover, drawn through any point of the circle as a point of tangency, is proved by the following theorem.
Theorem. If a line is perpendicular to the radius at its end lying on the circle, then this line is a tangent.
Let O (fig) be the center of some circle and OA some of its radius. Through its end A we draw MN ^ OA.
It is required to prove that the line MN is tangent, i.e. that this line has only one common point A with the circle.
Let us assume the opposite: let MN have another common point with the circle, for example B.
Then straight line OB would be a radius and therefore equal to OA.
But this cannot be, since if OA is perpendicular, then OB must be inclined to MN, and the inclined one is greater than the perpendicular.
Converse theorem. If a line is tangent to a circle, then the radius drawn to the point of tangency is perpendicular to it.
Let MN be the tangent to the circle, A the point of tangency, and O the center of the circle.
It is required to prove that OA^MN.
Let's assume the opposite, i.e. Let us assume that the perpendicular dropped from O to MN will not be OA, but some other line, for example, OB.
Let's take BC = AB and carry out OS.
Then OA and OS will be inclined, equally distant from the perpendicular OB, and therefore OS = OA.
It follows from this that the circle, taking into account our assumption, will have two common points with the line MN: A and C, i.e. MN will not be a tangent, but a secant, which contradicts the condition.
Consequence. Through any given point on a circle one can draw a tangent to this circle, and only one, since through this point one can draw a perpendicular, and only one, to the radius drawn into it.
Theorem. A tangent parallel to a chord divides the arc subtended by the chord in half at the point of contact.
Let straight line AB (fig.) touch the circle at point M and be parallel to chord CD.
We need to prove that ÈCM = ÈMD.
Drawing the diameter ME through the point of tangency, we obtain: EM ^ AB, and therefore EM ^ CB.
Therefore CM=MD.
Task. Through a given point draw a tangent to a given circle.
If a given point is on a circle, then draw a radius through it and a perpendicular straight line through the end of the radius. This line will be the desired tangent.
Let us consider the case when the point is given outside the circle.
Let it be required (Fig.) to draw a tangent to a circle with center O through point A.
To do this, from point A, as the center, we describe an arc with radius AO, and from point O, as the center, we intersect this arc at points B and C with a compass opening equal to the diameter of the given circle.
Having then drawn the chords OB and OS, we connect point A with points D and E, at which these chords intersect with the given circle.
Lines AD and AE are tangent to circle O.
Indeed, from the construction it is clear that the pipes AOB and AOC are isosceles (AO = AB = AC) with the bases OB and OS equal to the diameter of the circle O.
Since OD and OE are radii, then D is the middle of OB, and E is the middle of OS, which means AD and AE are medians drawn to the bases of isosceles pipes, and therefore perpendicular to these bases. If the lines DA and EA are perpendicular to the radii OD and OE, then they are tangent.
Consequence. Two tangents drawn from one point to a circle are equal and form equal angles with the straight line connecting this point to the center.
So AD=AE and ÐOAD = ÐOAE (Fig.), because rectangular tr-ki AOD and AOE, having a common hypotenuse AO and equal legs OD and OE (as radii), are equal.
Note that here the word “tangent” means the actual “tangent segment” from a given point to the point of contact.
Task. Draw a tangent to a given circle O parallel to a given straight line AB (Fig.).
We lower a perpendicular OS to AB from the center O and through the point D, at which this perpendicular intersects the circle, draw EF || AB.
The tangent we are looking for will be EF.
Indeed, since OS ^ AB and EF || AB, then EF ^ OD, and the line perpendicular to the radius at its end lying on the circle is a tangent.
Task. Draw a common tangent to two circles O and O 1 (Fig.).
Analysis. Let's assume that the problem is solved.
Let AB be the common tangent, A and B the points of tangency.
Obviously, if we find one of these points, for example, A, then we can easily find the other one.
Let us draw the radii OA and O 1 B. These radii, being perpendicular to the common tangent, are parallel to each other.
Therefore, if from O 1 we draw O 1 C || BA, then the pipeline OCO 1 will be rectangular at vertex C.
As a result, if we describe a circle from O as the center with radius OS, then it will touch the straight line O 1 C at point C.
The radius of this auxiliary circle is known: it is equal to OA – CA = OA - O 1 B, i.e. it is equal to the difference between the radii of these circles.
Construction. From the center O we describe a circle with a radius equal to the difference of these radii.
From O 1 we draw a tangent O 1 C to this circle (in the manner indicated in the previous problem).
Through the tangent point C we draw the radius OS and continue it until it meets the given circle at point A. Finally, from A we draw AB parallel to CO 1.
In exactly the same way we can construct another common tangent A 1 B 1 (Fig.). Direct lines AB and A 1 B 1 are called external common tangents.
You can spend two more internal tangents as follows:
Analysis. Let's assume that the problem is solved (Fig.). Let AB be the desired tangent.
Let us draw the radii OA and O 1 B to the tangent points A and B. Since these radii are both perpendicular to the common tangent, they are parallel to each other.
Therefore, if from O 1 we draw O 1 C || BA and continue OA to point C, then OS will be perpendicular to O 1 C.
As a result, the circle described by the radius OS from point O as the center will touch the straight line O 1 C at point C.
The radius of this auxiliary circle is known: it is equal to OA+AC = OA+O 1 B, i.e. it is equal to the sum of the radii of the given circles.
Construction. From O as the center, we describe a circle with a radius equal to the sum of these radii.
From O 1 we draw a tangent O 1 C to this circle.
We connect the point of contact C with O.
Finally, through point A, at which OS intersects the given circle, we draw AB = O 1 C.
In a similar way we can construct another internal tangent A 1 B 1.
General definition of tangent
Let a tangent AT and some secant AM be drawn through point A to a circle with a center (Fig.).
Let's rotate this secant around point A so that the other intersection point B moves closer and closer to A.
Then the perpendicular OD, lowered from the center to the secant, will approach the radius OA more and more, and the angle AOD may become less than any small angle.
The angle MAT formed by the secant and tangent is equal to the angle AOD (due to the perpendicularity of their sides).
Therefore, as point B approaches A indefinitely, angle MAT can also become arbitrarily small.
This is expressed in other words like this:
a tangent is the limiting position to which a secant drawn through a point of tangency tends when the second point of intersection approaches the point of tangency indefinitely.
This property is taken as the definition of a tangent when talking about any curve.
Thus, the tangent to the curve AB (Fig.) is the limiting position MT to which the secant MN tends when the intersection point P approaches M without limit.
Note that the tangent defined in this way can have more than one common point with the curve (as can be seen in Fig.).
Secant, tangent - all this could be heard hundreds of times in geometry lessons. But graduation from school is behind us, years pass, and all this knowledge is forgotten. What should you remember?
Essence
The term “tangent to a circle” is probably familiar to everyone. But it’s unlikely that everyone will be able to quickly formulate its definition. Meanwhile, a tangent is a straight line lying in the same plane as a circle that intersects it only at one point. There may be a huge number of them, but they all have the same properties, which will be discussed below. As you might guess, the point of tangency is the place where the circle and the straight line intersect. In each specific case there is only one, but if there are more of them, then it will be a secant.
History of discovery and study
The concept of a tangent appeared in ancient times. The construction of these straight lines, first to a circle, and then to ellipses, parabolas and hyperbolas using a ruler and compass, was carried out at the initial stages of the development of geometry. Of course, history has not preserved the name of the discoverer, but it is obvious that even at that time people were quite familiar with the properties of a tangent to a circle.
In modern times, interest in this phenomenon flared up again - a new round of study of this concept began, combined with the discovery of new curves. Thus, Galileo introduced the concept of a cycloid, and Fermat and Descartes constructed a tangent to it. As for circles, it seems that there are no secrets left for the ancients in this area.
Properties
The radius drawn to the intersection point will be This
the main, but not the only property that a tangent to a circle has. Another important feature includes two straight lines. So, through one point lying outside the circle, two tangents can be drawn, and their segments will be equal. There is another theorem on this topic, but it is rarely taught as part of a standard school course, although it is extremely convenient for solving some problems. It sounds like this. From one point located outside the circle, a tangent and a secant are drawn to it. The segments AB, AC and AD are formed. A is the intersection of lines, B is the point of tangency, C and D are intersections. In this case, the following equality will be valid: the length of the tangent to the circle, squared, will be equal to the product of the segments AC and AD.
There is an important corollary to the above. For each point on the circle you can construct a tangent, but only one. The proof of this is quite simple: theoretically dropping a perpendicular from the radius onto it, we find out that the formed triangle cannot exist. And this means that the tangent is the only one.
Construction
Among other problems in geometry there is a special category, as a rule, not
loved by pupils and students. To solve problems in this category, you only need a compass and a ruler. These are construction tasks. There are also ones for constructing a tangent.
So, given a circle and a point lying outside its boundaries. And it is necessary to draw a tangent through them. How to do this? First of all, you need to draw a segment between the center of the circle O and a given point. Then use a compass to divide it in half. To do this, you need to set a radius - a little more than half the distance between the center of the original circle and this point. After this, you need to build two intersecting arcs. Moreover, the radius of the compass does not need to be changed, and the center of each part of the circle will be the original point and O, respectively. The intersections of the arcs need to be connected, which will divide the segment in half. Set a radius on the compass equal to this distance. Next, construct another circle with the center at the intersection point. Both the original point and O will lie on it. In this case, there will be two more intersections with the circle given in the problem. They will be the points of contact for the initially specified point.
It was the construction of tangents to the circle that led to the birth
differential calculus. The first work on this topic was published by the famous German mathematician Leibniz. It provided for the possibility of finding maxima, minima and tangents regardless of fractional and irrational quantities. Well, now it is used for many other calculations.
In addition, the tangent to a circle is related to the geometric meaning of tangent. This is where its name comes from. Translated from Latin tangens means “tangent”. Thus, this concept is associated not only with geometry and differential calculus, but also with trigonometry.
Two circles
The tangent does not always affect only one figure. If a huge number of straight lines can be drawn to one circle, then why not vice versa? Can. But the task in this case becomes seriously complicated, because the tangent to two circles may not pass through any points, and the relative position of all these figures can be very
different.
Types and varieties
When we are talking about two circles and one or more straight lines, even if it is known that these are tangents, it is not immediately clear how all these figures are located in relation to each other. Based on this, several varieties are distinguished. Thus, circles may have one or two common points or not have them at all. In the first case they will intersect, and in the second they will touch. And here two varieties are distinguished. If one circle is, as it were, embedded in the second, then the tangency is called internal, if not, then external. You can understand the relative position of the figures not only based on the drawing, but also having information about the sum of their radii and the distance between their centers. If these two quantities are equal, then the circles touch. If the first one is greater, they intersect, and if it is less, then they do not have common points.
The same goes for straight lines. For any two circles that do not have common points, you can
construct four tangents. Two of them will intersect between the figures, they are called internal. A couple of others are external.
If we are talking about circles that have one common point, then the problem is greatly simplified. The fact is that, regardless of their relative position, in this case they will have only one tangent. And it will pass through the point of their intersection. So construction will not be difficult.
If the figures have two points of intersection, then a straight line can be constructed for them, tangent to the circle of both one and the other, but only external. The solution to this problem is similar to what will be discussed below.
Problem solving
Both the internal and external tangent to two circles are not so simple to construct, although this problem can be solved. The fact is that an auxiliary figure is used for this, so you have to come up with this method yourself
quite problematic. So, two circles with different radii and centers O1 and O2 are given. For them you need to construct two pairs of tangents.
First of all, you need to build an auxiliary one near the center of the larger circle. In this case, the difference between the radii of the two initial figures should be established on the compass. Tangents to the auxiliary circle are constructed from the center of the smaller circle. After this, perpendiculars are drawn from O1 and O2 to these lines until they intersect with the original figures. As follows from the basic property of the tangent, the required points on both circles are found. The problem is solved, at least the first part.
In order to construct internal tangents, you will have to solve practically
similar task. Again you will need an auxiliary figure, but this time its radius will be equal to the sum of the original ones. Tangents are constructed to it from the center of one of these circles. The further course of the solution can be understood from the previous example.
Tangent to a circle or even two or more is not such a difficult task. Of course, mathematicians have long stopped solving such problems manually and entrust the calculations to special programs. But you shouldn’t think that now you don’t have to be able to do it yourself, because to correctly formulate a task for a computer you need to do and understand a lot. Unfortunately, there are concerns that after the final transition to a test form of knowledge control, construction tasks will cause students more and more difficulties.
As for finding common tangents for a larger number of circles, this is not always possible, even if they lie in the same plane. But in some cases you can find such a straight line.
Examples from life
A common tangent to two circles often occurs in practice, although this is not always noticeable. Conveyors, block systems, pulley transmission belts, thread tension in a sewing machine, and even just a bicycle chain - all these are real-life examples. So you shouldn’t think that geometric problems remain only in theory: in engineering, physics, construction and many other fields they find practical application.
Direct ( MN), having only one common point with the circle ( A), called tangent to the circle.
The common point is called in this case point of contact.
Possibility of existence tangent, and, moreover, drawn through any point circle, as a point of tangency, is proven as follows theorem.
Let it be required to carry out circle with center O tangent through the point A. To do this from the point A, as from the center, we describe arc radius A.O., and from the point O, as the center, we intersect this arc at the points B And WITH a compass solution equal to the diameter of the given circle.
After spending then chords O.B. And OS, connect the dot A with dots D And E, at which these chords intersect with a given circle. Direct AD And A.E. - tangents to a circle O. Indeed, from the construction it is clear that triangles AOB And AOC isosceles(AO = AB = AC) with bases O.B. And OS, equal to the diameter of the circle O.
Because O.D. And O.E.- radii, then D - middle O.B., A E- middle OS, Means AD And A.E. - medians, drawn to the bases of isosceles triangles, and therefore perpendicular to these bases. If straight D.A. And E.A. perpendicular to the radii O.D. And O.E., then they - tangents.
Consequence.
Two tangents drawn from one point to a circle are equal and form equal angles with the straight line connecting this point to the center.
So AD=AE and ∠ OAD = ∠OAE because right triangles AOD And AOE, having a common hypotenuse A.O. and equal legs O.D. And O.E.(as radii), are equal. Note that here the word “tangent” actually means “ tangent segment” from a given point to the point of contact.
\[(\Large(\text(Central and inscribed angles)))\]
Definitions
A central angle is an angle whose vertex lies at the center of the circle.
An inscribed angle is an angle whose vertex lies on a circle.
The degree measure of an arc of a circle is the degree measure of the central angle that subtends it.
Theorem
The degree measure of an inscribed angle is equal to half the degree measure of the arc on which it rests.
Proof
We will carry out the proof in two stages: first, we will prove the validity of the statement for the case when one of the sides of the inscribed angle contains a diameter. Let point \(B\) be the vertex of the inscribed angle \(ABC\) and \(BC\) be the diameter of the circle:
Triangle \(AOB\) is isosceles, \(AO = OB\) , \(\angle AOC\) is external, then \(\angle AOC = \angle OAB + \angle ABO = 2\angle ABC\), where \(\angle ABC = 0.5\cdot\angle AOC = 0.5\cdot\buildrel\smile\over(AC)\).
Now consider an arbitrary inscribed angle \(ABC\) . Let us draw the diameter of the circle \(BD\) from the vertex of the inscribed angle. There are two possible cases:
1) the diameter cuts the angle into two angles \(\angle ABD, \angle CBD\) (for each of which the theorem is true as proven above, therefore it is also true for the original angle, which is the sum of these two and therefore equal to half the sum of the arcs to which they rest, that is, equal to half the arc on which it rests). Rice. 1.
2) the diameter did not cut the angle into two angles, then we have two more new inscribed angles \(\angle ABD, \angle CBD\), whose side contains the diameter, therefore, the theorem is true for them, then it is also true for the original angle (which is equal to the difference of these two angles, which means it is equal to the half-difference of the arcs on which they rest, that is, equal to half the arc on which it rests). Rice. 2.
Consequences
1. Inscribed angles subtending the same arc are equal.
2. An inscribed angle subtended by a semicircle is a right angle.
3. An inscribed angle is equal to half the central angle subtended by the same arc.
\[(\Large(\text(Tangent to the circle)))\]
Definitions
There are three types of relative positions of a line and a circle:
1) straight line \(a\) intersects the circle at two points. Such a line is called a secant line. In this case, the distance \(d\) from the center of the circle to the straight line is less than the radius \(R\) of the circle (Fig. 3).
2) straight line \(b\) intersects the circle at one point. Such a line is called a tangent, and their common point \(B\) is called the point of tangency. In this case \(d=R\) (Fig. 4).
Theorem
1. A tangent to a circle is perpendicular to the radius drawn to the point of tangency.
2. If a line passes through the end of the radius of a circle and is perpendicular to this radius, then it is tangent to the circle.
Consequence
The tangent segments drawn from one point to a circle are equal.
Proof
Let us draw two tangents \(KA\) and \(KB\) to the circle from the point \(K\):
This means that \(OA\perp KA, OB\perp KB\) are like radii. Right triangles \(\triangle KAO\) and \(\triangle KBO\) are equal in leg and hypotenuse, therefore, \(KA=KB\) .
Consequence
The center of the circle \(O\) lies on the bisector of the angle \(AKB\) formed by two tangents drawn from the same point \(K\) .
\[(\Large(\text(Theorems related to angles)))\]
Theorem on the angle between secants
The angle between two secants drawn from the same point is equal to the half-difference in degree measures of the larger and smaller arcs they cut.
Proof
Let \(M\) be the point from which two secants are drawn as shown in the figure:
Let's show that \(\angle DMB = \dfrac(1)(2)(\buildrel\smile\over(BD) - \buildrel\smile\over(CA))\).
\(\angle DAB\) is the external angle of the triangle \(MAD\), then \(\angle DAB = \angle DMB + \angle MDA\), where \(\angle DMB = \angle DAB - \angle MDA\), but the angles \(\angle DAB\) and \(\angle MDA\) are inscribed, then \(\angle DMB = \angle DAB - \angle MDA = \frac(1)(2)\buildrel\smile\over(BD) - \frac(1)(2)\buildrel\smile\over(CA) = \frac(1)(2)(\buildrel\smile\over(BD) - \buildrel\smile\over(CA))\), which was what needed to be proven.
Theorem on the angle between intersecting chords
The angle between two intersecting chords is equal to half the sum of the degree measures of the arcs they cut: \[\angle CMD=\dfrac12\left(\buildrel\smile\over(AB)+\buildrel\smile\over(CD)\right)\]
Proof
\(\angle BMA = \angle CMD\) as vertical.
From triangle \(AMD\) : \(\angle AMD = 180^\circ - \angle BDA - \angle CAD = 180^\circ - \frac12\buildrel\smile\over(AB) - \frac12\buildrel\smile\over(CD)\).
But \(\angle AMD = 180^\circ - \angle CMD\), from which we conclude that \[\angle CMD = \frac12\cdot\buildrel\smile\over(AB) + \frac12\cdot\buildrel\smile\over(CD) = \frac12(\buildrel\smile\over(AB) + \buildrel\ smile\over(CD)).\]
Theorem on the angle between a chord and a tangent
The angle between the tangent and the chord passing through the point of tangency is equal to half the degree measure of the arc subtended by the chord.
Proof
Let the straight line \(a\) touch the circle at the point \(A\), \(AB\) is the chord of this circle, \(O\) is its center. Let the line containing \(OB\) intersect \(a\) at the point \(M\) . Let's prove that \(\angle BAM = \frac12\cdot \buildrel\smile\over(AB)\).
Let's denote \(\angle OAB = \alpha\) . Since \(OA\) and \(OB\) are radii, then \(OA = OB\) and \(\angle OBA = \angle OAB = \alpha\). Thus, \(\buildrel\smile\over(AB) = \angle AOB = 180^\circ - 2\alpha = 2(90^\circ - \alpha)\).
Since \(OA\) is the radius drawn to the tangent point, then \(OA\perp a\), that is, \(\angle OAM = 90^\circ\), therefore, \(\angle BAM = 90^\circ - \angle OAB = 90^\circ - \alpha = \frac12\cdot\buildrel\smile\over(AB)\).
Theorem on arcs subtended by equal chords
Equal chords subtend equal arcs smaller than semicircles.
And vice versa: equal arcs are subtended by equal chords.
Proof
1) Let \(AB=CD\) . Let us prove that the smaller semicircles of the arc .
On three sides, therefore, \(\angle AOB=\angle COD\) . But because \(\angle AOB, \angle COD\) - central angles supported by arcs \(\buildrel\smile\over(AB), \buildrel\smile\over(CD)\) accordingly, then \(\buildrel\smile\over(AB)=\buildrel\smile\over(CD)\).
2) If \(\buildrel\smile\over(AB)=\buildrel\smile\over(CD)\), That \(\triangle AOB=\triangle COD\) on two sides \(AO=BO=CO=DO\) and the angle between them \(\angle AOB=\angle COD\) . Therefore, and \(AB=CD\) .
Theorem
If the radius bisects the chord, then it is perpendicular to it.
The converse is also true: if the radius is perpendicular to the chord, then at the point of intersection it bisects it.
Proof
1) Let \(AN=NB\) . Let us prove that \(OQ\perp AB\) .
Consider \(\triangle AOB\) : it is isosceles, because \(OA=OB\) – radii of the circle. Because \(ON\) is the median drawn to the base, then it is also the height, therefore, \(ON\perp AB\) .
2) Let \(OQ\perp AB\) . Let us prove that \(AN=NB\) .
Similarly, \(\triangle AOB\) is isosceles, \(ON\) is the height, therefore, \(ON\) is the median. Therefore, \(AN=NB\) .
\[(\Large(\text(Theorems related to the lengths of segments)))\]
Theorem on the product of chord segments
If two chords of a circle intersect, then the product of the segments of one chord is equal to the product of the segments of the other chord.
Proof
Let the chords \(AB\) and \(CD\) intersect at the point \(E\) .
Consider the triangles \(ADE\) and \(CBE\) . In these triangles, angles \(1\) and \(2\) are equal, since they are inscribed and rest on the same arc \(BD\), and angles \(3\) and \(4\) are equal as vertical. Triangles \(ADE\) and \(CBE\) are similar (based on the first criterion of similarity of triangles).
Then \(\dfrac(AE)(EC) = \dfrac(DE)(BE)\), from which \(AE\cdot BE = CE\cdot DE\) .
Tangent and secant theorem
The square of a tangent segment is equal to the product of a secant and its outer part.
Proof
Let the tangent pass through the point \(M\) and touch the circle at the point \(A\) . Let the secant pass through the point \(M\) and intersect the circle at the points \(B\) and \(C\) so that \(MB< MC\) . Покажем, что \(MB\cdot MC = MA^2\) .
Consider the triangles \(MBA\) and \(MCA\) : \(\angle M\) is common, \(\angle BCA = 0.5\cdot\buildrel\smile\over(AB)\). According to the theorem about the angle between a tangent and a secant, \(\angle BAM = 0.5\cdot\buildrel\smile\over(AB) = \angle BCA\). Thus, triangles \(MBA\) and \(MCA\) are similar at two angles.
From the similarity of triangles \(MBA\) and \(MCA\) we have: \(\dfrac(MB)(MA) = \dfrac(MA)(MC)\), which is equivalent to \(MB\cdot MC = MA^2\) .
Consequence
The product of a secant drawn from the point \(O\) by its external part does not depend on the choice of the secant drawn from the point \(O\) .
Definition. A tangent to a circle is a straight line in a plane that has exactly one common point with the circle.
Here are a couple of examples:
Circle with center O touches a straight line l at the point A
From anywhere M Exactly two tangents can be drawn outside the circle 
Difference between tangent l, secant B.C. and straight m, which has no common points with a circle We could end here, but practice shows that it is not enough to simply memorize the definition - you need to learn to see tangents in drawings, know their properties, and, in addition, practice properly in applying these properties by solving real problems. We will do all of this today.
Basic properties of tangents
In order to solve any problem, you need to know four key properties. Two of them are described in any reference book/textbook, but the last two are somehow forgotten about, but in vain.
1. Tangent segments drawn from one point are equal
A little higher we already talked about two tangents drawn from one point M. So:
Tangent segments to a circle drawn from one point are equal.
Segments A.M. And B.M. equal 2. The tangent is perpendicular to the radius drawn to the point of tangency
Let's look at the picture above again. Let's draw the radii O.A. And O.B., after which we find that the angles OAM And O.B.M.- straight.
The radius drawn to the point of contact is perpendicular to the tangent.
This fact can be used without proof in any problem:
Radii drawn to the tangent point are perpendicular to the tangents By the way, note: if you draw a segment OM, then we get two equal triangles: OAM And O.B.M..
3. Relationship between tangent and secant
But this is a more serious fact, and most schoolchildren do not know it. Consider a tangent and a secant that pass through the same common point M. Naturally, the secant will give us two segments: inside the circle (segment B.C.- it is also called the chord) and outside (it is also called the outer part M.C.).
The product of the entire secant and its external part is equal to the square of the tangent segment
Relationship between secant and tangent 4. Angle between tangent and chord
An even more advanced fact that is often used to solve complex problems. I highly recommend taking it into service.
The angle between the tangent and the chord is equal to the inscribed angle subtended by this chord.
Where does the point come from? B? In real problems, it usually “pops up” somewhere in the condition. Therefore, it is important to learn to recognize this configuration in drawings.
Sometimes it does matter :)
