How to construct a conjugation of a right angle. Making a drawing of a part with mates. Conjugation of circles (arcs) with a straight line

Pairing.

Pairing is a smooth transition from one line to another.

Conjugation of intersecting lines by an arc of a circle of a given radius.

The problem is reduced to drawing a circle tangent to both given straight lines.

Option 1.

We draw auxiliary lines parallel to the given ones at a distance R from the given ones.

The point of intersection of these lines will be the center O conjugation arcs. Perpendiculars dropped from the center O to

the given straight lines will determine the tangent points K and K 1 .

Option 2.

The construction is the same.

Pairings. Construction of conjugation of lines.

Option 3.

If you want to draw a circle so that it touches three intersecting straight lines, then in this case

The radius cannot be specified by the conditions of the problem. Center O circle is at the intersection bisector corners

AT and FROM. The radius of the circle is the perpendicular dropped from the center O to any of the 3 given lines

Lines.

Pairings. Construction of line conjugations.

Construction of an external conjugation of a given circle with a given straight arc of a given radius R 1 .

From the center O of this circle we draw an arc of an auxiliary circle with a radius R+R 1 .

We draw a straight line parallel to the given one at a distance R1.

The intersection of the straight line and the construction arc will give the center point of the fillet arc About 1 .

Point of contact of arcs To lies on the line OO 1 .

Point of contact between arc and line K 1 lies at the intersection of the perpendicular from the point O 1 to the line with the arc.

Pairings. Construction of external conjugation of a circle with a straight line.

Construction of an internal conjugation of a given circle with a given straight arc of a given radius R 1 .

From the center O of this circle we draw an auxiliary circle with a radius R- R 1 .

Pairings. Construction of internal conjugation of a circle with a straight line.

Construction of conjugation of two given circles by an arc of a given radius R 3 .

External touch.

From the center of the circle About 1 R1+R3.

From the center of the circle About 2 describe the arc of the auxiliary circle with radius R2+R3.

intersection arcs of auxiliary circles will give a point About 3, which is the center of the conjugation arc

touch points K 1 and K 2 are on the lines O 1 O 3 and O 2 O 3.

Inner touch

From the center of the circle About 1 describe the arc of the auxiliary circle with radius R 3 -R 1 .

From the center of the circle About 2 describe the arc of the auxiliary circle with radius R 3 - R 2 .

intersection

(circles with radius R 3) .

Pairings. Conjugation of two circles by an arc.

External and internal touch.

Given two circles with centers O 1 and O 2 with radii r 1 and r 2 . It is necessary to draw a circle of a given

Radius R so as to provide internal contact with one circle, and external contact with the other.

From the center of the circle About 1 describe the arc of the auxiliary circle with radius R-r 1 .

From the center of the circle About 2 describe the arc of the auxiliary circle with radius R+r 2 .

intersectionarcs of auxiliary circles will give a point that is the center of the conjugation arc

(circles with radius R) .

Pairings. Conjugation of two circles by an arc.

Construction of a circle passing through given point And tangent to the given circle

at a given point B.

Finding the midpoint of a straight line AB. Through the middle of the line AB draw a perpendicular. Continuation intersection

OB and perpendicular lines give a point About 1 . About 1 - the center of the desired circle with a radius R = O 1 B = O 1 A.

Pairings. Internal tangency of circle and arc.

Constructing a conjugation of a circle with a straight line at a given point A on a straight line.

From a given point A of the line LM, we restore the perpendicular to the straight line LM. On the continuation

Perpendicular set aside a segment AB. AB = R. We connect the point B with the center of the circle O 1 straight line.

From point A we draw a straight line parallel to BO 1 until it intersects with a circle. Let's get a point To- point

Touch. Connect the point K with the center of the circle O 1 . Let's extend the lines O 1 K and AB to the intersection. Let's get a point

About 2, which is the center of the conjugation arc with radius O 2 A \u003d O 2 K.

Pairings. Conjugation of a circle with a straight line at a given point.

Constructing a conjugation of a circle with a straight line at a point A given on the circle.

External touch.

We spend tangent to a circle through a point BUT. The intersection of a tangent with a straight line LM will give a point AT.

Dividing the corner in half

About 1. About 1 O 1 A \u003d O 1 K.

Internal touch.

We spend tangent to a circle through a point BUT. The intersection of the tangent line with the LM line will give a point AT.

Dividing the corner, formed by the tangent and straight line LM , in half. The intersection of the bisector of the angle and

Extending the radius OA will give a point About 1. About 1 - O 1 A \u003d O 1 K.

Pairings. The conjugation of a circle with a straight line at a given point on the circle.

Construction of a conjugation of two non-concentric arcs of circles by an arc of a given radius.

Draw from the center of the arc About 1 auxiliary arc with radius R 1 -R 3 . Draw from the center of the arc O 2 auxiliary

Arc radius R2+R3. The intersection of the arcs will give a point Oh. Oh- the center of the arc of conjugation with the radius R3. touch points

K 1 and K 2 lie on the lines OO 1 and OO 2.

Pairings. Pairing 2 non-concentric arcs of circles with an arc.

Construction of a curved curve by selecting arcs.

By selecting the centers of arcs coinciding with sections of the curve, you can draw any curved curve with a compass.

In order for the arcs to smoothly transition from one to another, it is necessary that the points of their conjugation (tangency)

They were on straight lines connecting the centers of these arcs.

The sequence of constructions.

We select the center 1 arcs of an arbitrary section ab.

On the continuation first radius select the center 2 plot arc radius b.c.

On the continuation second radius select the center 3 plot arc radius cd etc.

So we build the whole curve.

Pairings. Selection of arcs.

Construction of conjugation of two parallel lines by two arcs.

Points defined on straight parallel lines BUT and AT connect with a line AB.

Choose on a straight line AB arbitrary point M.

We divide the segments AM and VM in half.

We restore perpendiculars in the middle of the segments.

At points A and B, given lines, we restore the perpendiculars to the lines.

intersection relevant perpendiculars will give points About 1 and About 2.

About 1 center of arc of conjugation with radius O 1 A \u003d O 1 M.

About 2 center of arc of conjugation with radius O 2 V \u003d O 2 M.

If the point M choose on middle lines AB, then radii conjugation arcs will are equal.

Touching arcs at a point M located on the line About 1 About 2.

Pairings. Conjugation of parallel lines by two arcs.

Pairing Center- a point equidistant from the mating lines. And the common point for these lines is called conjugation point .

The construction of conjugations is performed using a compass.

The following types of pairing are possible:

1) conjugation of intersecting lines using an arc of a given radius R (rounding corners);

2) conjugation of a circular arc and a straight line using an arc of a given radius R;

3) conjugation of arcs of circles of radii R 1 and R 2 by a straight line;

4) conjugation of arcs of two circles of radii R 1 and R 2 by an arc of a given radius R (external, internal and mixed conjugation).

With external mating, the centers of the mating arcs of radius R 1 and R 2 lie outside the mating arc of radius R. With internal mating, the centers of the mating arcs lie inside the mating arc of radius R. With mixed mating, the center of one of the mating arcs lies inside the mating arc of radius R, and the center of the other mating arc - outside it.

In table. 1 shows the construction and gives brief explanations for the construction of simple conjugations.

PairingsTable 1

An example of simple mates	Graphic construction of mates	Brief explanation for the construction
1. Conjugation of intersecting lines using an arc of a given radius R.		Draw straight lines parallel to the sides of the angle at a distance R. From a point O mutual intersection of these lines, lowering the perpendiculars to the sides of the angle, we get the conjugation points 1 and 2 . Radius R draw an arc.
2. Conjugation of a circular arc and a straight line using an arc of a given radius R.		On distance R draw a line parallel to a given line, and from the center O 1 with a radius R+R 1- an arc of a circle. Dot O- the center of the conjugation arc. Point 2 we get on the perpendicular drawn from point O to a given straight line, and point 1 - on a straight line OO 1 .
3. Conjugation of arcs of two circles of radii R1 and R2 straight line.		From point O 1 draw a circle with radius R 1 - R2. The segment O 1 O 2 is divided in half and from the point O 3 draw an arc with a radius of 0.5 O 1 O 2 . Connect points O 1 and O 2 with a point BUT. From point O 2, drop the perpendicular to the line AO 2, points 1.2 - pairing points.

Table 1 continued

4. Conjugation of arcs of two circles of radii R1 and R2 arc of a given radius R(external pairing).		From centers O 1 and O 2 draw arcs of radii R+R 1 and R + R 2 . O 1 and O 2 with point O. Points 1 and 2 are junction points.
5. Conjugation of arcs of two circles of radii R1 and R2 arc of a given radius R(internal pairing).		From centers O 1 and O 2 draw arcs of radii R-R1 and R-R2. We get a point O- the center of the conjugation arc. connect the dots O 1 and O 2 with point O until the intersection with the given circles. points 1 and 2- junction points.
6. Conjugation of arcs of two circles of radii R1 and R2 arc of a given radius R(mixed conjugation).		From the centers O 1 and O 2 draw arcs of radii R- R 1 and R + R 2 . We get point O - the center of the conjugation arc. connect the dots O 1 and O 2 with point O until the intersection with the given circles. points 1and 2- junction points.

curved curves

These are curved lines, in which the curvature continuously changes on each of their elements. Curved curves cannot be drawn with a compass, they are constructed from a series of points. When drawing a curve, the resulting series of points is connected along a pattern, so it is called a curved line. The accuracy of building a curved curve increases with an increase in the number of intermediate points on a curve section.

The curved curves include the so-called flat sections of the cone - ellipse, parabola, hyperbola, which are obtained as a result of the section of a circular cone by a plane. Such curves were considered when studying the course "Descriptive Geometry". Curves also include involute, sinusoid, spiral of Archimedes, cycloidal curves.

Ellipse- the locus of points, the sum of the distances of which to two fixed points (foci) is a constant value.

The most widely used method of constructing an ellipse along the given semiaxes AB and CD. When constructing, two concentric circles are drawn, the diameters of which are equal to the given axes of the ellipse. To build 12 points of an ellipse, the circles are divided into 12 equal parts and the resulting points are connected to the center.

On fig. 15 shows the construction of six points of the upper half of the ellipse; the lower half is drawn in the same way.

Involute- is the trajectory of a circle point formed by its deployment and straightening (circle development).

The construction of an involute according to a given diameter of a circle is shown in fig. 16. The circle is divided into eight equal parts. From points 1,2,3 draw tangents to the circle, directed in one direction. On the last tangent, the involute step is set equal to the circumference

(2 pR), and the resulting segment is also divided into 8 equal parts. Putting one part on the first tangent, two parts on the second, three parts on the third, etc., we get the involute points.

Cycloid curves- flat curved lines described by a point belonging to a circle rolling without slipping along a straight line or circle. If at the same time the circle rolls in a straight line, then the point describes a curve called a cycloid.

The construction of a cycloid according to a given circle diameter d is shown in Fig.17.

Rice. 17

A circle and a segment of length 2pR are divided into 12 equal parts. Draw a straight line through the center of the circle parallel to the line segment. From the points of division of the segment to the straight line, perpendiculars are drawn. At the points of their intersection with the straight line, we get O 1, O 2, O 3, etc. are the centers of the rolling circle.

From these centers we describe arcs of radius R. Through the division points of the circle we draw straight lines parallel to the straight line connecting the centers of the circles. At the intersection of the straight line passing through point 1 with the arc described from the center O1, there is one of the points of the cycloid; through point 2 with another from the center O2 - another point, etc.

If the circle rolls along another circle, being inside it (along the concave part), then the point describes a curve called hypocycloid. If a circle rolls along another circle, being outside it (along the convex part), then the point describes a curve called epicycloid.

The construction of a hypocycloid and an epicycloid is similar, but instead of a segment of length 2pR, an arc of the guide circle is taken.

The construction of an epicycloid according to a given radius of the movable and fixed circles is shown in Fig.18. Angle α, which is calculated by the formula

α = 180°(2r/R), and the circle of radius R is divided into eight equal parts. An arc of a circle of radius R + r is drawn and from points О 1 , О 2 , О 3 .. - a circle of radius r.

The construction of the hypocycloid by the given radii of the moving and fixed circles is shown in Fig.19. The angle α, which is calculated, and the circle of radius R are divided into eight equal parts. An arc of a circle with radius R - r is drawn and from points O 1, O 2, O 3 ... - a circle with radius r.

Parabola- this is the locus of points equidistant from a fixed point - the focus F and a fixed line - the directrix, perpendicular to the axis of symmetry of the parabola. The construction of a parabola according to a given segment OO \u003d AB and a chord CD is shown in Fig. 20

Direct OE and OS are divided into the same number of equal parts. Further construction is clear from the drawing.

Hyperbola- the locus of points, the difference in the distances of which from two fixed points (foci) - is a constant value. Represents two open, symmetrically located branches.

The constant points of the hyperbola F 1 and F 2 are foci, and the distance between them is called focal. The line segments connecting the points of the curve with the foci are called radius vectors. A hyperbola has two mutually perpendicular axes - real and imaginary. The lines passing through the center of intersection of the axes are called asymptotes.

The construction of a hyperbola according to a given focal length F 1 F 2 and the angle α between the asymptotes is shown in Fig.21. An axis is drawn on which the focal length is plotted, which is halved by point O. A circle of radius 0.5F 1 F 2 is drawn through point O until it intersects at points C, D, E, K. Connecting points C with D and E with K, one obtains points A and B are the vertices of the hyperbola. From point F 1 to the left, arbitrary points 1, 2, 3 are marked ... the distances between which should increase as they move away from the focus. From focal points F 1 and F 2 with radii R=B4 and r=A4, arcs are drawn to mutual intersection. Intersection points 4 are points of the hyperbola. The rest of the points are constructed in a similar way.

sinusoid- a flat curve expressing the law of the change in the sine of the angle depending on the change in the magnitude of the angle.

The construction of a sinusoid for a given circle diameter d is shown

in fig. 22.

To build it, divide given circle into 12 equal parts; a segment equal to the length of a given circle (2pR) is divided into the same number of equal parts. Drawing horizontal and vertical straight lines through the division points, they find the sinusoid points at their intersection.

Spiral of Archimedes - e then a plane curve, described by a point, which rotates uniformly around a given center and at the same time uniformly moves away from it.

The construction of the Archimedes spiral for a given circle diameter D is shown in Fig.23.

The circumference and the radius of the circle are divided into 12 equal parts. Further construction is visible from the drawing.

When constructing conjugations and curved curves, one has to resort to the simplest geometric constructions - such as dividing a circle or a straight line into several equal parts, dividing an angle and a segment in half, building perpendiculars, bisectors, etc. All these constructions were studied in the discipline "Drawing" school course and are therefore not discussed in detail in this manual.

1.5 Guidelines for the implementation

In the general case, the construction of a conjugation of a circle m of radius R 1 and a straight line l with a circle of radius R (Fig. 30, a, b) is carried out as follows:

- at a distance R parallel to l we draw l '(GM to the straight line);

- with the center at the point O 1 we draw m '(GM to the circle), with a radius equal to the sum of R and R 1 or a radius equal to the difference R and R 1 ;

– the point О of the intersection of l’ and m’ is the conjugation center;

- we drop the perpendicular from O to the line l. We get the junction point A;

- draw a straight line through O and O 1 and mark the conjugation point B of its intersection with the circle m;

- with the center at point O with radius R between points A and B, we draw an arc of conjugation.

Rice. 30. Conjugation of a straight line with a circle

Conjugation of two circles

When building external pairing two circles m 1 and m 2 by an arc of a given radius R (Fig. 31) the center of the mating arc - point O - is determined by the intersection of two geometric places m 1 ' and m 2 ' - auxiliary circles of radii R + R 1 and R + R 2, drawn respectively from the centers of the conjugated circles, i.e. from points O 1 and O 2. Conjugation points A and B are defined as the points of intersection of the given circles with straight lines OO 1 and OO 2.

Internal pairing arcs of radii R 1 and R 2 with an arc of radius R is shown in fig. 32.

Rice. 31. External pairing of two circles

Rice. 32. Internal conjugation of two circles

To determine the center O of the conjugation arc, we draw auxiliary arcs m 1 ’and m 2 ’ from the points O 1 and O 2 - two geometric places - with radii R–R 1 and R–R 2. The point of intersection of these arcs is the center of conjugation. From the point O through the points O 1 and O 2 we draw straight lines to the intersection with the circles m 1 and m 2 and we get the conjugation points A and B. Between these points an arc of the conjugation circle of radius R is drawn with the center at the point O.

At mixed conjugation(Fig. 33) the conjugation center O is determined at the intersection of two geometric places - auxiliary circles of radii R + R 1 and R–R 2, drawn from the centers O 1 and O 2, respectively. Conjugation points A and B lie at the intersection of the lines of centers OO 1 and OO 2 with arcs of given circles.

Rice. 33. Construction of a mixed conjugation of two circles

Construction of tangent lines

The construction of tangents to circles is based on the fact that the tangent line is perpendicular to the radius of the circle drawn to the point of contact.

Construction of a tangent to a circle from a point A lying outside the circle (Fig. 34). The segment OA connecting the given point A with the center O of the circle is divided in half and from the resulting point O 1, as from the center, we describe the auxiliary circle with radius O 1 A. The auxiliary circle intersects the given one at point B, which is the point of contact. The line AB will be tangent to the circle, because angle ABO is right, as inscribed in an auxiliary circle and based on its diameter.

Construction of a tangent to two circles. A tangent to two circles can be external if both circles are located on the same side of it, and internal if the circles are located on different sides of the tangent.

Rice. 34. Construction of a tangent to a circle

To build an external tangent to circles of radii R 1 and R 2 (Fig. 35), we proceed as follows:

one). from the center O 2 of the larger circle we draw an auxiliary circle with a radius R 2 -R 1;

2). segment O 1 O 2 is divided in half;

3). with the center O 3 we draw an auxiliary circle with a radius O 3 O 2;

four). mark the intersection points of two auxiliary circles - M and N;

5). draw straight lines through the point O 2 and the obtained points until they intersect with a circle of radius R 2 . We get points B and D;

6). from the center O 1 we draw straight lines O 1 A and O 1 C, respectively, parallel to O 2 B and O 2 D until they intersect with a circle of radius R 1 at points A and C.

Straight lines AB and CD are the desired external tangents to two circles.

Rice. 35. Construction of an external tangent to two circles

Construction of an internal tangent to two circles of radii R 1 and R 2 (Fig. 36).

Rice. 36. Construction of an internal tangent to two circles

From the center of one of the circles, for example from O 1, we draw an auxiliary circle with a radius R 1 + R 2. We divide the segment O 1 O 2 in half and from the obtained point O 3 we draw a second auxiliary circle with a radius O 1 O 3. We connect the points M and N of the intersection of the auxiliary circles with straight lines with the center O 1 and at their intersection with a circle of radius R 1 we get the points of contact A and C. From the point O 2 we draw a straight line parallel to O 1 A and we get the point of contact B on the circle R 2. The point D is constructed similarly. The lines AB and CD are the required internal tangents to the two circles.

Lesson number 23.

Pairings

Show multiple parts that have fillets.

Examining the details, we see that in their design often one surface passes into another. Usually these transitions are made smooth, which increases the strength of the parts and makes them more convenient to work with.

In the drawing, the surfaces are depicted by lines that also smoothly transition from one to another.

Such a smooth transition from one line (surface) to another line (surface) is called conjugation.

When constructing a conjugation, it is necessary to determine the boundary where one line ends and another begins, i.e. find the transition point in the drawing, which is called conjugation point or touch point .

Conjugation problems can be conditionally divided into 3 groups.

First group of tasks includes tasks for constructing mates, where straight lines are involved. This can be a direct touch of a line and a circle, conjugation of two lines by an arc of a given radius, as well as drawing a tangent line to two circles.

Construct a circle tangent to a straight line.

Construction of a circle tangent to a straight line , is related to finding the point of contact and the center of the circle.

Given a horizontal line AB , it is required to construct a circle with a radius R tangent to the given line (Fig. 1).

The touch point is chosen arbitrarily.

Since the tangent point is not specified, the circle of radius R can touch this line at any point. There are many such circles. The centers of these circles ( O 1 , O 2 etc.) will be at the same distance from the given straight line, i.e. on a line parallel to a given line AB at a distance equal to the radius of a given circle (Fig. 1). Let's call this line center line .

Draw a line of centers parallel to a straight line AB on distance R . Since the center of the tangent circle is not set, we take any point on the center line, for example, the point O.

Before drawing a tangent circle, the point of tangency must be determined. The point of contact will lie on the perpendicular dropped from the point O directly AB . At the intersection of a perpendicular with a line AB get a point TO, which will be the point of contact. From the center O radius R from the point To let's draw a circle. Problem solved.

Write in your notebooks in a cage following rules:

If a straight line is involved in the conjugation, then:

1)

the center of a circle tangent to a straight line lies on a straight line (line of centers) drawn parallel to a given straight line, at a distance equal to the radius of the given circle;

2) the point of contact lies on a perpendicular drawn from the center of the circle to a given straight line.

Conjugation of two lines.

On a plane, two straight lines can be parallel or at an angle to each other.

To construct a conjugation of two lines, it is necessary to draw a circle tangent to these two lines.

Open your workbooks to page 31.

Consider the conjugation of two non-parallel lines.

Two non-parallel lines are located at an angle to each other, which can be straight, obtuse or acute. When making drawings of parts, such corners often need to be rounded with an arc of a given radius (Fig. 1). The rounding of corners in the drawing is nothing more than the conjugation of two non-parallel straight lines with an arc of a circle of a given radius. To perform pairing, you need to find the center of the pairing arc and the pairing points.

It is known that if a straight line is involved in the conjugation, then the center of the conjugation arc is located on the center line, which is drawn parallel to the given straight line at a distance equal to the radius R conjugation arcs.

Since the angle is formed by two straight lines, two lines of centers are drawn parallel to each straight line at a distance equal to the radius R conjugation arcs. The point of their intersection will be the center of the conjugation arc.

To find conjugation points from a point O drop perpendiculars to the given lines and get conjugation points To and To 1 . Knowing the points and the center of conjugation, from the point O radius R conduct an arc of conjugation. When tracing a drawing, first trace the arc, and then the tangent lines.

When building a conjugation right angle drawing the line of centers is simplified, since the sides of the angle are mutually perpendicular. From the top of the corner lay segments equal to the radius R conjugation arcs, and through the obtained points To and To 1 , which will be the points of contact, draw two lines of centers parallel to the sides of the corner. They will be both center lines and perpendiculars that define the junction points. To and To 1 (p. 31, fig. 1).

Page 31, task 4. Conjugation of two parallel lines.

To build a conjugation of two parallel lines, it is necessary to draw an arc of a circle tangent to these lines (Fig. 3).

Fig.3

The radius of this circle will be equal to half the distance between the given lines. Since the tangent point is not given, there are many such circles that can be drawn. Their centers will be on a straight line drawn parallel to the given straight lines at a distance, half distances between them. This straight line will be the line of centers.

touch points ( To 1 and To 2 ) lie on the perpendicular dropped from the center of the tangent circle to the given lines (Fig. 3a). Since the center of the tangent circle is not specified, the perpendicular is drawn arbitrarily. Line segment QC 1 are divided in half (Fig. 3b), a straight line is drawn through the points of intersection of the serifs parallel to the given lines, on which the centers of the circles tangent to the given parallel lines will be located, i.e. this line will be the line of centers. Putting the leg of the compass on a point O , draw an arc of conjugation (Fig. 3c) from the point To to the point To 1 .

Construction of lines tangent to circles

(R.T. p.33).

Exercise 1. Draw a line tangent to a circle through a point BUT lying on the circle.

From a point O draw a straight line OB through a point BUT . From a point BUT Draw a circle with any radius. At the intersection with a straight line received points 1 and 2. From these points with any radius we draw arcs until they intersect each other at points C and D . From a point C or D draw a line through a point BUT .

It will be tangent to the circle, since the tangent is always perpendicular to the radius drawn to the tangent point.

Task 2.

This construction is similar to the construction of a perpendicular to a straight line through a given point, which can be done using two squares.

First a square 1 is placed so that its hypotenuse coincides with the points O and A . Then to square 1 a square is applied 2 , which will be the guide, i.e. along which the square will move 1 . Then a square 1 attach another leg to the square 2. Then we roll the square 1 by square 2 until the hypotenuse coincides with the point A . And we draw a line tangent to the circle through a point A .

Task 3. Draw a line tangent to a circle through a point not on the circle.

Given a circle with a radiusR and dot BUT , not lying on the circle, it is required to draw from the pointBUT a straight line tangent to the given circle in its upper part. To do this, you need to find the point of contact. We know that the tangent point lies on the perpendicular drawn from the center of the circle to the tangent line. Therefore, the tangent and the perpendicular form a right angle.

Knowing that every angle inscribed in a circle and based on its diameter is a right angle, connecting the pointsBUT and O , take a segmentJSC for the diameter of the circumscribed circle. At the intersection of the circumscribed circle and the circle of radiusR will be the vertex of the right angle (pointTo ). Line segment JSC divide in half with a compass, get a pointO 1 (Fig. 4, b).

From the center O 1 radius equal to the segmentJSC 1 , draw a circle, get pointsTo and To 1 at the intersection with a circle of radiusR (Fig. 4, c).

Since you need to draw only one tangent to the top of the circle, choose desired point touch. This point will be the pointTo . Point To connect with dotsBUT and O , we get a right angle, which relies on the diameterJSC circumscribed circle with radiusR 1 . Dot To - the vertex of this angle (Fig. 4, d), segmentsOK and AK - sides of a right angle, therefore, a pointTo will be the desired point of contact, and the straight lineAK - the desired tangent.

Fig.4

Drawing a line tangent to two circles.

Given two circles with radii R and R 1 , it is required to construct a tangent to them. There are two cases of contact: external and internal.

With external tangency, the tangent line is on the same side of the circles and does not intersect the segment connecting the centers of these circles.

With internal tangency, the tangent line is on different sides of the circles and intersects the segment connecting the centers of the circles.

Page 33. Task 5. Draw a line tangent to two circles. The touch is external.

First of all, you need to find the points of contact. It is known that they must lie on perpendiculars drawn from the centers of the circles ( O and O 1 ) to the tangent.

From a point O draw a circle with a radius R - R 1 , since the touch is external.

Divide the distance OO 1 in half and draw a circle with a radius R =OO 2 =O 1 O 2

This circle intersects a circle with radius R - R 1 at the point TO. We connect this point with O 1 .

From a point O through a point To draw a straight line until it intersects with a circle of radius R . got a point To 1 - the first point of contact.

From a point O 1 draw a line parallel QC 1 , until it intersects with a circle of radius R 1 . Got a second touch point To 2 . Connecting the dots To 1 and To 2 . This is the tangent to the two circles.

Task 6. Draw a line tangent to two circles. The touch is internal.

The construction is similar, only with internal contact the radius of the auxiliary circle drawn from the point O is equal to the sum of the radii of the circles R + R 1 .

The second group of pairing problems includes tasks that involve only circles and arcs. A smooth transition from one circle to another can occur either directly by touch, or through the third element - the arc of a circle.

The tangency of two circles can be external (PT: p.32, fig.3) or internal (PT: p.32, fig.4).

Task 3 (page 32)

When two circles touch externally, the distance between the centers of these circles will be equal to the sum of their radii.

From a point O radius R + R C let's make an arc. From a point O 1 radius R 1 + R C O FROM - center of conjugation.

Connecting the dots O and O 1 with the center of conjugation O FROM . On the circles got points of contact (conjugation).

From a point O FROM mate radius R C 30 connecting points of contact.

Task 4 (page 32)

When two circles touch internally, one of the tangent circles is inside the other circle, and the distance between the centers of these circles will be equal to the difference of their radii.

From a point O radius ( R C – R ) let's make an arc. From a point O 1 radius ( R C – R 1 ) draw an arc until it intersects with the first arc. got a point O FROM - center of conjugation.

Pairing Center O FROM connect with dots O and O 1 with and extend the straight line further.

On the circles got points of contact (conjugation).

From a point O FROM mate radius R C 60 connecting points of contact.

The third group of pairing problems includes tasks for conjugation of a straight line and an arc of a circle with an arc of a given radius.

Performing such a task, they solve, as it were, two problems: drawing a tangent arc to a straight line and a tangent arc to a circle. The touch in this case can be both external and internal.

RT: page 32. Task 1. Conjugation of a circle and a straight line. The touch is external. R C 20 .

Given a straight line and a circle with a radius R , it is required to construct a conjugation by an arc of radius R C 20 .

Since a straight line is involved in the mate, the center of the mate arc is on a straight line drawn parallel to the given line at a distance equal to the mate radius R C 20 . Therefore, parallel to the given straight line at a distance of 20 mm, we draw another straight line.

And the center of the conjugation arc, when the two circles touch externally, is located on the circle of radius, equal to the sum radii R and R C . Therefore, from the point O radius ( R + R C O FROM

Then we find the points of contact. The first point of contact is a perpendicular dropped from the center of the mate to the given line. We find the second junction point by connecting the junction center O FROM and the center of the circle R . The tangent point will lie at the first intersection with the circle, since the tangency is external.

Then from the point O FROM radius R C 20 connect the points of intersection.

RT: page 32. Task 2. Conjugation of a circle and a straight line. The touch is internal. R C 60 .

Draw a line of centers parallel to the given straight line at a distance of 60 mm. From a point O radius ( R With - R ) we draw an arc to the intersection with a new straight line (line of centers). Let's get a point O FROM , which is the center of conjugation.

From O FROM draw a line through the center of the circle O and a perpendicular to a given line. We get two points of contact. And then from the center of the pairing with a radius of 60 mm we connect the points of contact.

Pairing is a smooth transition from one line to another. A smooth transition can be done both with the help of circular lines
(arcs of circles), and with the help of curved curves (arcs of an ellipse, parabola or hyperbola). We will consider only cases of conjugations with the help of arcs of circles. From all the variety of conjugations various lines the following main types of conjugations can be distinguished: conjugation of two differently located straight lines using an arc of a circle, conjugation of a straight line with an arc of a circle, construction of a common tangent to two circles, conjugation of two circles of a third. Any type of pairing should be performed in the following sequence:

- find the center of the conjugation arc,

- find connection points,

- a conjugation arc is drawn with a given radius.

Different kinds mates are shown in Table 2:

table 2

Graphic construction of mates	Brief explanation for the construction
Conjugation of intersecting lines by an arc of a given radius
	Draw straight lines parallel to the sides of the corner at a distance R. From the point O, the mutual intersection of these lines, dropping the perpendiculars to the sides of the corner, we get conjugation points 1 and 2. With a radius R, draw an arc of conjugation between points 1 and 2.
Conjugation of a circle and a straight line using an arc of a given radius
	At a distance R, draw a straight line parallel to a given straight line, and from the center O 1 with radius R + R 1 - an arc of a circle. Point O is the center of the conjugation arc. We get point 2 on the perpendicular dropped from point O to a given straight line, and point 1 - at the intersection of line OO 1 and a circle of radius R.

Continuation of table 2

Conjugation of arcs of two circles by a straight line
	From point O, draw an auxiliary circle with radius R-R 1. Divide the segment OO 1 in half and from point O 2 draw a circle with a radius of 0.5 OO 1. This circle intersects the auxiliary at point K 0. By connecting the point K 0 with the point O 1 we get the direction of the common tangent. The tangent points K and K 1 are found at the intersection of perpendiculars from points O and O 1 with given circles.
Conjugation of arcs of two circles with an arc of a given radius (external conjugation)
	From the centers O 1 and O 2, draw arcs of radii R + R 1 and R + R 2. When these arcs intersect, we get a point O - the center of the conjugation arc. Connect points O 1 and O 2 with point O. Points K and K 1 are conjugation points. Between points K and K 1 draw an arc of conjugation with a radius R.

Continuation of table 2

Conjugation of arcs of two circles with an arc of a given radius (internal conjugation)
	Draw arcs from the centers O 1 and O 2 radii R-R 1 and R-R 2 . At the intersection of these arcs, we obtain a point O - the center of the conjugation arc. Connect points O 1 and O 2 with point O until they intersect with the given circles. Points K and K 1 are conjugation points. Between points K and K 1 with radius R we draw an arc of conjugation.
Conjugation of arcs of two circles with an arc of a given radius (mixed conjugation)
	From the centers O 1 and O 2, draw arcs of radii R-R 1 and R + R 2. At the intersection of these arcs, we obtain a point O - the center of the conjugation arc. We connect the points O 1 and O 2 with the point O until they intersect with the given circles. Points 1 and 2 are conjugation points. Between points 1 and 2 with radius R we draw an arc of conjugation.