Properties of a tangent. Tangent line Tangent rule

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 (that’s what they call it - 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 :)

\[(\Large(\text(Central and inscribed angles)))\]

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 \(O\) .

Points x_0\in\mathbb(R), and is differentiable in it: f \in \mathcal(D)(x_0). Tangent line to the graph of a function f at the point x_0 is called the graph of a linear function given by the equation y = f(x_0) + f"(x_0)(x-x_0),\quad x\in \mathbb(R).

  • If the function f has at point x_0 infinite derivative f"(x_0) = \pm \infty, then the tangent line at this point is the vertical line given by the equation x = x_0.

    • Comment

    It follows directly from the definition that the graph of a tangent line passes through the point (x_0,f(x_0)). Corner \alpha between the tangent to the curve and the Ox axis satisfies the equation

    \operatorname(tg)\,\alpha = f"(x_0)= k,

    Where \operatorname(tg) denotes tangent, and \operatorname (k)- tangent slope coefficient. Derivative at a point x_0 equal to the slope of the tangent to the graph of the function y = f(x) at this point.

    Tangent as the limit position of a secant

    Let f\colon U(x_0) \to \R And x_1 \in U(x_0). Then a straight line passing through the points (x_0,f(x_0)) And (x_1,f(x_1)) given by the equation

    y = f(x_0) + \frac(f(x_1) - f(x_0))(x_1 - x_0)(x-x_0).

    This line passes through the point (x_0,f(x_0)) for anyone x_1\in U(x_0), and its angle of inclination \alpha(x_1) satisfies the equation

    \operatorname(tg)\,\alpha(x_1) = \frac(f(x_1) - f(x_0))(x_1 - x_0).

    Due to the existence of the derivative function f at the point x_0, going to the limit at x_1 \to x_0, we find that there is a limit

    \lim\limits_(x_1 \to x_0) \operatorname(tg)\,\alpha(x_1) = f"(x_0),

    and due to the continuity of the arctangent and the limiting angle

    \alpha = \operatorname(arctg)\,f"(x_0).

    Line passing through a point (x_0,f(x_0)) and having a maximum angle of inclination that satisfies \operatorname(tg)\,\alpha = f"(x_0), is given by the tangent equation:

    y = f(x_0) + f"(x_0)(x-x_0).

    Tangent to a circle

    A straight line that has one common point with a circle and lies in the same plane with it is called a tangent to the circle.

    Properties

    1. A tangent to a circle is perpendicular to the radius drawn to the point of tangency.
    2. Tangent segments to a circle drawn from one point are equal and make equal angles with a straight line passing through this point and the center of the circle.
    3. The length of a tangent segment drawn to a circle of unit radius, taken between the point of tangency and the point of intersection of the tangent with a ray drawn from the center of the circle, is the tangent of the angle between this ray and the direction from the center of the circle to the point of tangency. "Tangent" from lat. tangents- “tangent”.

    Variations and generalizations

    One-sided half-tangents

    • If there is a right derivative f"_+(x_0)< \infty, That right half-tangent to the graph of the function f at the point x_0 called a ray
    y = f(x_0) + f"_+(x_0)(x - x_0),\quad x\geqslant x_0.
    • If there is a left derivative f"_-(x_0)< \infty, That left half-tangent to the graph of the function f at the point x_0 called a ray
    y = f(x_0) + f"_-(x_0)(x - x_0),\quad x \leqslant x_0.
    • If there is an infinite right derivative f"_+(x_0) = +\infty\; (-\infty), f at the point x_0 called a ray
    x = x_0,\; y\geqslant f(x_0)\; (y \leqslant f(x_0)).
    • If there is an infinite left derivative f"_-(x_0) = +\infty\; (-\infty), then the right half-tangent to the graph of the function f at the point x_0 called a ray
    x = x_0,\; y \leqslant f(x_0)\; (y \geqslant f(x_0)).

    see also

    • Normal, binormal

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    Literature

    • Toponogov V. A. Differential geometry of curves and surfaces. - Fizmatkniga, 2012. - ISBN 9785891552135.
    • // Encyclopedic Dictionary of Brockhaus and Efron: in 86 volumes (82 volumes and 4 additional). - St. Petersburg. , 1890-1907.

    An excerpt characterizing the tangent line

    - In places! - the young officer shouted at the soldiers gathered around Pierre. This young officer, apparently, was fulfilling his position for the first or second time and therefore treated both the soldiers and the commander with particular clarity and formality.
    The rolling fire of cannons and rifles intensified throughout the entire field, especially to the left, where Bagration’s flashes were, but because of the smoke of the shots, it was impossible to see almost anything from the place where Pierre was. Moreover, observing the seemingly family (separated from all others) circle of people who were on the battery absorbed all of Pierre’s attention. His first unconscious joyful excitement, produced by the sight and sounds of the battlefield, was now replaced, especially after the sight of this lonely soldier lying in the meadow, by another feeling. Now sitting on the slope of the ditch, he observed the faces surrounding him.
    By ten o'clock twenty people had already been carried away from the battery; two guns were broken, shells hit the battery more and more often, and long-range bullets flew in, buzzing and whistling. But the people who were at the battery did not seem to notice this; Cheerful talk and jokes were heard from all sides.
    - Chinenka! - the soldier shouted at the approaching grenade flying with a whistle. - Not here! To the infantry! – another added with laughter, noticing that the grenade flew over and hit the covering ranks.
    - What, friend? - another soldier laughed at the man who crouched under the flying cannonball.
    Several soldiers gathered at the rampart, looking at what was happening ahead.
    “And they took off the chain, you see, they went back,” they said, pointing across the shaft.
    “Mind your job,” the old non-commissioned officer shouted at them. “We’ve gone back, so it’s time to go back.” - And the non-commissioned officer, taking one of the soldiers by the shoulder, pushed him with his knee. There was laughter.
    - Roll towards the fifth gun! - they shouted from one side.
    “At once, more amicably, in the burlatsky style,” the cheerful cries of those changing the gun were heard.
    “Oh, I almost knocked off our master’s hat,” the red-faced joker laughed at Pierre, showing his teeth. “Eh, clumsy,” he added reproachfully to the cannonball that hit the wheel and the man’s leg.
    - Come on, you foxes! - another laughed at the bending militiamen entering the battery behind the wounded man.
    - Isn’t the porridge tasty? Oh, the crows, they slaughtered! - they shouted at the militia, who hesitated in front of the soldier with a severed leg.
    “Something else, kid,” they mimicked the men. – They don’t like passion.
    Pierre noticed how after each cannonball that hit, after each loss, the general revival flared up more and more.
    As if from an approaching thundercloud, more and more often, lighter and brighter, lightning of a hidden, flaring fire flashed on the faces of all these people (as if in rebuff to what was happening).
    Pierre did not look forward to the battlefield and was not interested in knowing what was happening there: he was completely absorbed in the contemplation of this increasingly flaring fire, which in the same way (he felt) was flaring up in his soul.
    At ten o'clock the infantry soldiers who were in front of the battery in the bushes and along the Kamenka River retreated. From the battery it was visible how they ran back past it, carrying the wounded on their guns. Some general with his retinue entered the mound and, after talking with the colonel, looked angrily at Pierre, went down again, ordering the infantry cover stationed behind the battery to lie down so as to be less exposed to shots. Following this, in the ranks of the infantry, to the right of the battery, a drum and command shouts were heard, and from the battery it was visible how the ranks of the infantry moved forward.
    Pierre looked through the shaft. One face in particular caught his eye. It was an officer who, with a pale young face, walked backwards, carrying a lowered sword, and looked around uneasily.
    The rows of infantry soldiers disappeared into the smoke, and their prolonged screams and frequent gunfire could be heard. A few minutes later, crowds of wounded and stretchers passed from there. Shells began to hit the battery even more often. Several people lay uncleaned. The soldiers moved more busily and more animatedly around the guns. Nobody paid attention to Pierre anymore. Once or twice they shouted at him angrily for being on the road. The senior officer, with a frowning face, moved with large, fast steps from one gun to another. The young officer, flushed even more, commanded the soldiers even more diligently. The soldiers fired, turned, loaded, and did their job with tense panache. They bounced as they walked, as if on springs.

    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 in combination 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, using a compass, 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.

    Same with 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.(like radii) are equal. Note that here the word “tangent” actually means “ tangent segment” from a given point to the point of contact.