Another way to find the center (for example, of turned products) - using a special tool, a “centre finder” - is based on the properties of the so-called. tangent lines. A tangent to a circle is any straight line that, at the point of meeting the circle, is perpendicular to the radius drawn to this point. For example, to hell. 174 straight AB, CD And E.F.– tangents to a circle ACE. Points A, C, E are called "touch points". The peculiarity of a tangent line is that it has a circle with only one common point. Indeed, if the tangent AB(Fig. 175) was with a circle, besides this there is another common point, for example, WITH, then, connecting it to the center, we would get an isosceles triangle SOA with two right angles SA, and this, we know, is impossible (why?).

We encounter lines tangent to a circle quite often in practical life. A rope thrown over a block takes in its tense parts the position of tangent lines to the circle of the block. The belts of hoists (combinations of several blocks, Fig. 176) are located along the line of common tangents to the circumference of the wheels. The transmission belts of the pulleys also occupy the position of common tangents to the circles of the pulleys of “external” tangents in the so-called. open transmission and “internal” - in closed transmission.

How to draw a tangent to it through a given point outside the circle? In other words: like through a dot A(drawing 177) draw a straight line AB to angle ABO was it straight? This is done as follows. Connect A with center ABOUT(drawing 178). The straight line is divided in half and around its middle IN, as a center, describe a circle with a radius IN. In other words, on OA build a circle as on a diameter. Intersection points WITH And D both circles are connected to A straight lines: these will be tangents.

To verify this, let’s draw from the center to the points WITH And D auxiliary lines OS And OD. Angles WASP And ODA- straight, since they are inscribed in a semicircle. And this means that OS And O.D.– tangents to the circle.

Considering our construction, we see, among other things, that from each point outside the circle we can draw two tangents to it. It is easy to verify that both of these tangents are of the same length, i.e., that A.C.= AD. Indeed, period ABOUT equally distant from the sides of the angle A; Means OA is an equidivisor, and therefore triangles OAS And OAD equal ( SUS).

Along the way, we established that the straight line that bisects the angle between both tangents passes through the center of the circle. This is the basis for the design of the device for finding the center of turned products - the center of the finder (Fig. 179). It consists of two lines AB And AC, fixed at an angle, and the third ruler BD, the edge of which BD bisects the angle between the edges

the first two lines. The device is applied to the round product so that the edges of the rulers adjacent to it AB And Sun came into contact with the circumference of the product. In this case, the edges will have only one common point with the circle, so the edge of the ruler must, according to the now indicated property of tangents, pass through the center of the circle. Having drawn the diameter of a circle on the product using a ruler, apply the center finder to the product in a different position and draw a different diameter. The desired center will be at the intersection of both diameters.

If you need to draw a common tangent to two circles, that is, draw a straight line that would touch two circles at the same time, then proceed as follows. Near the center of one circle, for example, about IN(Fig. 180), describe an auxiliary circle with a radius equal to the difference between the radii of both circles. Then from the point A draw tangents AC And AD to this auxiliary circle. From points A And IN draw straight lines perpendicular to AC And AD, until they intersect with the given circles at points E, F, H And G. Straight lines connecting E With F, G With H, there will be common tangents to these circles, since they are perpendicular to the radii AE, CF, AG And D.H..

In addition to the two tangents that have just been drawn and which are called external, it is also possible to draw two other tangents, located like hell. 181 (internal tangents). To perform this construction, describe around the center of one of these circles - for example, around IN– an auxiliary circle with a radius equal to the sum of the radii of both circles. From the point A draw tangents to this auxiliary circle. Readers will be able to find out the further course of construction themselves.

Repeat questions

What is a tangent called? How many common points do the tangent and the circle have in common? – How to draw a tangent to a circle through a point lying outside the circle? – How many such tangents can be drawn? – What is a centrifuge? – What is its device based on? – How to draw a common tangent to two circles? - How many tangents are there?

In this chapter we will return to one of the basic geometric shapes - the circle. Various theorems related to circles will be proven, including theorems about circles inscribed in a triangle, quadrilateral, and circumscribed circles around these figures. In addition, three statements will be proven about the remarkable points of a triangle - the point of intersection of the triangle's bisectors, the point of intersection of its altitudes, and the point of intersection of the perpendicular bisectors to the sides of the triangle. The first two statements were formulated back in 7th grade, and now we can prove them.

Let's find out how many common points a straight line and a circle can have, depending on their relative position. It is clear that if a line passes through the center of a circle, then it intersects the circle at two points - the ends of the diameter lying on this line.

Let the straight line p not pass through the center O of a circle of radius r. Let us draw a perpendicular OH to the straight line p and denote by the letter d the length of this perpendicular, that is, the distance from the center of this circle to the straight line (Fig. 211).

Rice. 211

Let's study the relative position of the straight line and the circle depending on the relationship between d and r. There are three possible cases.

1) d< r. На прямой р от точки Н отложим два отрезка НА и НВ, длины которых равны (рис. 211, а). По теореме Пифагора

Consequently, points A and B lie on the circle and, therefore, are common points of the straight line p and the given circle.

Let us prove that the straight line p and the given circle have no other common points. Let us assume that they have one more common point C. Then the median OD of the isosceles triangle O AC drawn to the base AC is the height of this triangle, therefore OD ⊥ p. Segments OD and OH do not coincide, since midpoint D of segment AC does not coincide with point H - the midpoint of segment AB. We found that from point O two perpendiculars (segments OH and OD) were drawn to straight line p, which is impossible.

So, if the distance from the center of the circle to the straight line is less than the radius of the circle (d< r), то прямая и окружность имеют две общие точки . In this case, the straight line is called a secant with respect to the circle.

2) d = r. In this case OH = r, i.e. point H lies on the circle and, therefore, is the common point of the line and the circle (Fig. 211.6). The straight line p and the circle have no other common points, since for any point M of the straight line p, different from the point H, OM > OH = r (the inclined OM is greater than the perpendicular OH), and, therefore, the point M does not lie on the circle.

So, if the distance from the center of the circle to the straight line is equal to the radius of the circle, then the straight line and the circle have only one common point.

3) d > r. In this case, OH > r, therefore, for any point M of the line r OM ≥ OH > r (Fig. 211, c). Therefore, point M does not lie on the circle.

So, if the distance from the center of the circle to the straight line is greater than the radius of the circle, then the straight line and the circle have no common points.

Tangent to a circle

We have proven that a line and a circle can have one or two common points and may not have any common points.

A straight line that has only one common point with a circle is called a tangent to the circle, and their common point is called the tangent point of the line and the circle. In Figure 212, straight line p is tangent to a circle with center O, A is the point of tangency.

Let us prove a theorem about the property of a tangent to a circle.

Theorem

Proof

Let p be the tangent to the circle with center O, A the point of tangency (see Fig. 212). Let us prove that the tangent p is perpendicular to the radius OA.

Rice. 212

Let's assume that this is not the case. Then the radius OA is inclined to the straight line r. Since the perpendicular drawn from point O to straight line p is less than inclined OA, the distance from the center O of the circle to straight line p is less than the radius. Consequently, the straight line p and the circle have two common points. But this contradicts the condition: straight line p is tangent.

Thus, straight line p is perpendicular to the radius OA. The theorem has been proven.

Consider two tangents to a circle with center O, passing through point A and touching the circle at points B and C (Fig. 213). Let's call the segments AB and AC tangent segments drawn from a point A. They have the following property:

Rice. 213

To prove this statement, let's turn to Figure 213. According to the theorem on the tangent property, angles 1 and 2 are right angles, therefore triangles ABO and ACO are right-angled. They are equal because they have a common hypotenuse OA and equal legs OB and OS. Therefore, AB = AC and ∠3 = ∠4, which is what needed to be proven.

Let us now prove the theorem converse to the theorem about the tangent property (tangent property).

Theorem

Proof

From the conditions of the theorem it follows that this radius is a perpendicular drawn from the center of the circle to the given line. Therefore, the distance from the center of the circle to the straight line is equal to the radius, and, therefore, the straight line and the circle have only one common point. But this means that this line is tangent to the circle. The theorem has been proven.

The solution to problems involving constructing a tangent line is based on this theorem. Let's solve one of these problems.

Task

Through a given point A of a circle with center O, draw a tangent to this circle.

Solution

Let's draw a straight line O A, and then construct a straight line p passing through point A perpendicular to the straight line O A. According to the tangent criterion, the straight line p is the desired tangent.

Tasks

631. Let d be the distance from the center of a circle of radius r to a straight line r. What is the relative position of the straight line r and the circle if: a) r = 16 cm, d = 12 cm; b) r = 5 cm, d = 4.2 cm; c) r = 7.2 dm, (2 = 3.7 dm; d) r = 8 cm, d = 1.2 dm; e) r = 5 cm, d = 50 mm?

632. The distance from point A to the center of the circle is less than the radius of the circle. Prove that any line passing through point A is a secant with respect to the given circle.

633. Given a square O ABC, the side of which is 6 cm, and a circle with a center at point O of radius 5 cm. Which of the lines OA, AB, BC and AC are secant with respect to this circle?

634. The radius OM of a circle with center O divides the chord AB in half. Prove that the tangent drawn through the point M is parallel to the chord AB.

635. A tangent and a chord equal to the radius of the circle are drawn through point A of the circle. Find the angle between them.

636. Two tangents are drawn through the ends of the chord AB, equal to the radius of the circle, intersecting at point C. Find the angle AC B.

637. The angle between the diameter AB and the chord AC is 30°. A tangent is drawn through point C and intersects line AB at point D. Prove that triangle ACD is isosceles.

638. Line AB touches a circle with center O of radius r at point B. Find AB if OA = 2 cm and r = 1.5 cm.

639. Line AB touches a circle with center O of radius r at point B. Find AB if ∠AOB = 60° and r = 12 cm.

640. Given a circle with center O of radius 4.5 cm and point A. Two tangents to the circle are drawn through point A. Find the angle between them if OA = 9 cm.

641. Segments AB and AC are tangent segments to a circle with center O, drawn from point A. Find angle BAC if the midpoint of segment AO lies on the circle.

642. In Figure 213 OB = 3cm, CM. = 6 cm. Find AB, AC, ∠3 and ∠4.

643. Lines AB and AC touch a circle with center O at points B and C. Find BC if ∠OAB = 30°, AB = 5 cm.

644. Straight lines MA and MB touch a circle with center O at points A and B. Point C is symmetrical to point O relative to point B. Prove that ∠AMC = 3∠BMC.

645. From the ends of the diameter AB of a given circle, perpendiculars AA 1 and BB 1 are drawn to the tangent, which is not perpendicular to the diameter AB. Prove that the point of tangency is the midpoint of the segment A 1 B 1 .

646. In triangle ABC, angle B is right. Prove that: a) straight line BC is tangent to a circle with center A of radius AB; b) straight line AB is tangent to a circle with center C of radius CB; c) straight line AC is not tangent to circles with center B and radii BA and BC.

647. Segment AN is a perpendicular drawn from point A to a straight line passing through the center O of a circle of radius 3 cm. Is straight line AN tangent to the circle if: a) CM. = 5 cm, AN = 4 cm; b) ∠HAO = 45°, CM = 4 cm; c) ∠HAO = 30°, O A = 6 cm?

648. Construct a tangent to a circle with center O: a) parallel to the given line; b) perpendicular to a given line.

Answers to problems

Lessons on the COMPASS program.

Lesson #12. Constructing circles in Compass 3D.
Circles tangent to curves, a circle based on two points.

Compass 3D has several ways to construct tangent circles:

• circle tangent to the 1st curve;
• circle tangent to 2 curves;
• circle tangent to 3 curves;

To construct a circle tangent to the curve, press the button "Circle tangent to 1 curve" in the compact panel, or in the top menu, press the commands sequentially "Tools" - "Geometry" - "Circles" - "Circle tangent to 1 curve."

Using the cursor, we first indicate the curve through which the circle will pass, then specify the 1st and 2nd points of this circle (the coordinates of the points can be entered in the property panel).

Phantoms of all possible circle options will be displayed on the screen. Using the cursor, select those that we need and fix them by clicking the “Create object” button. We complete the construction by clicking the “Abort command” button.

Before specifying the second point, you can enter a radius or diameter value in the corresponding field on the property panel. Such a circle will not always be constructed. This depends on the given radius or diameter. The impossibility of construction will be indicated by the disappearance of the phantom after entering the radius value.

If the center point of the circle is known, it can also be set in the properties panel.

To construct a circle tangent to two curves, press the button "Circle tangent to 2 curves" in a compact panel. Or in the top menu, press the commands sequentially "Tools" - "Geometry" - "Circles" - "Circle tangent to 2 curves".

Using the cursor, we indicate the objects that the circle should touch. Phantoms of all possible construction options will be displayed on the screen.

If the position of a point belonging to the circle is known, then it must be specified using the cursor, or the coordinates must be entered in the property panel. You can also enter radius or diameter values ​​in the properties panel. To complete the construction, select the desired phantom and press the buttons successively "Create object" And "Abort command".

To construct a circle tangent to three curves, press the button "Circle tangent to 3 curves" in a compact panel. Or in the top menu, press the commands sequentially "Tools" - "Geometry" - "Circles" - "Circle tangent to 3 curves."

The constructions are similar to the previous ones, so do them yourself, the result is shown in the figure below.

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.

State budgetary educational institution

Gymnasium No. 000

Design work in geometry.

Eight ways to construct a tangent to a circle.

9 biological-chemical class

Scientific director: ,

Deputy Director for Academic Affairs,

mathematics teacher.

Moscow 2012

Introduction

Chapter 1. …………………………………………………………………………………4

Conclusion

Introduction

The highest manifestation of the spirit is the mind.

The highest manifestation of reason is geometry.

The geometry cell is a triangle. He also

inexhaustible, like the universe. The circle is the soul of geometry.

Know the circle and you not only know the soul

geometry, but also elevate your soul.

Claudius Ptolemy
Task.

Construct a tangent to a circle with center O and radius R passing through point A lying outside the circle

Chapter 1.

Construction of a tangent to a circle that does not require justification based on the theory of parallel lines.

https://pandia.ru/text/78/156/images/image003_18.gif" width="17" height="16 src=">ABO = 90°. For a circle (O; r) OB - radius. OB AB, therefore, AB is a tangent according to the tangent property.

Similarly, AC is a tangent to a circle.

Construction No. 1 is based on the fact that the tangent of a circle is perpendicular to the radius drawn to the point of contact.

For a straight line there is only one point of contact with a circle.

Only one perpendicular line can be drawn through a given point on a line.

Construction No. 2.

https://pandia.ru/text/78/156/images/image003_18.gif" width="17" height="16"> ABO = 90°

5. OB – radius, ABO = 90°, therefore, AB – tangent by attribute.

6. Similarly, in the isosceles triangle AON AC is the tangent (ACO = 90°, OS is the radius)

7. So, AB and AC are tangents

Formation No. 3

https://pandia.ru/text/78/156/images/image003_18.gif" width="17" height="16">ORM = OVA = 90° (as corresponding angles in equal triangles), therefore, AB – tangent based on tangent.

4. Similarly, AC is a tangent

Construction №4

https://pandia.ru/text/78/156/images/image008_9.jpg" align="left" width="330" height="743 src=">

Construction No. 6.

Construction:

2. I will draw an arbitrary straight line through point A intersecting the circle (O, r) at points M and N.

6. AB and BC are the required tangents.

Proof:

1. Since triangles PQN and PQM are inscribed in a circle and side PQ is the diameter of the circle, then these triangles are right-angled.

2. In triangle PQL, segments PM and QN are heights intersecting at point K, therefore KL is the third height..gif" width="17" height="16 src=">.gif" width="17" height="16 src =">AQS =AMS = 180° - https://pandia.ru/text/78/156/images/image003_18.gif" width="17" height="16">PQN = β, then |AQ| = |AS|ctg β. Therefore, |PA| : |AQ| = ctg α: ctg β (2).

5. Comparing (1) and (2) I get |PD| : |PA| = |DQ| : |AQ|, or

(|OD| + R)(|OA| - R)=(R -|OD|)(|OA| + R).

After opening the brackets and simplifying, I find that |OD|·|OA|=R².

5. From the relation |OD|·|OA|=R² it follows that |OD|:R=R: |OA|, that is, triangles ODB and OBA are similar..gif" width="17" height="16"> OBA = 90°. Therefore, straight line AB is the desired tangent, which was what needed to be proven.

Construction No. 6.

Construction:

1. I will construct a circle (A; |OA|).

2. I will find a compass opening equal to 2R, for which I will select point S on the circle (O; R) and plot three arcs containing 60º each: SP=PQ=QT=60°. Points S and T are diametrically opposed.

3. I build a circle (O; ST) intersecting w 1What kind of circle is this? at points M and N.

4. Now I’ll build the middle of the MO. To do this, I construct circles (O; OM) and (M; MO), and then for points M and O we find diametrically opposite points U and V on them.

6. Finally, I will construct a circle (K; KM) and (L; LM), intersecting at the desired point B - the middle of MO.

Proof:

Triangles KMV and UMK are isosceles and similar. Therefore, from the fact that KM = 0.5 MU, it follows that MB = 0.5 MK = 0.5 R. So, point B is the desired point of contact. Similarly, you can find the point of contact C.

Chapter 3.

Construction of a tangent to a circle based on the properties of secants and bisectors.

Formation No. 7

https://pandia.ru/text/78/156/images/image011_7.jpg" align="left" width="440" height="514 src="> Formation No. 8

Construction:

1. Construct a circle (A;AP) intersecting straight line AP at point D.

2. Construct a circle w on the diameter QD

3. I will intersect it with a perpendicular to the straight line AP at point A and get points M and N.

Proof:

It is obvious that AM²=AN²=AD·AQ=AP·AQ. Then the circle (A;AM) intersects (O;R) at the tangent points B and C. AB and AC are the required tangents.