Segments of tangents to a circle drawn. Reference materials: chord, secant, tangent, theorems

The concept of a tangent to a circle

A circle has three possible mutual arrangements relatively straight:

If the distance from the center of the circle to the straight line is less than the radius, then the straight line has two points of intersection with the circle.

If the distance from the center of the circle to the straight line is equal to the radius, then the straight line has two points of intersection with the circle.

If the distance from the center of the circle to the straight line is greater than the radius, then the straight line has two points of intersection with the circle.

Let us now introduce the concept of a tangent line to a circle.

Definition 1

A tangent to a circle is a line that has one intersection point with it.

The common point of the circle and the tangent is called the point of tangency (Figure 1).

Figure 1. Tangent to a circle

Theorems related to the concept of a tangent to a circle

Theorem 1

Tangent property theorem: a tangent to a circle is perpendicular to the radius drawn to the point of tangency.

Proof.

Consider a circle with center $O$. Let us draw tangent $a$ at point $A$. $OA=r$ (Fig. 2).

Let us prove that $a\bot r$

We will prove the theorem by contradiction. Suppose that the tangent $a$ is not perpendicular to the radius of the circle.

Figure 2. Illustration of Theorem 1

That is, $OA$ is inclined to the tangent. Since the perpendicular to the straight line $a$ is always less than the inclined one to the same straight line, the distance from the center of the circle to the straight line is less than the radius. As we know, in this case the straight line has two points of intersection with the circle. Which contradicts the definition of a tangent.

Therefore, the tangent is perpendicular to the radius of the circle.

The theorem has been proven.

Theorem 2

Converse of the tangent property theorem: If a line passing through the end of the radius of a circle is perpendicular to the radius, then this line is tangent to this circle.

Proof.

According to the conditions of the problem, we have that the radius is a perpendicular drawn from the center of the circle to a given straight line. Therefore, the distance from the center of the circle to the straight line is equal to the length of the radius. As we know, in this case the circle has only one point of intersection with this line. By Definition 1 we find that this line is tangent to the circle.

The theorem has been proven.

Theorem 3

Segments of tangents 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.

Proof.

Let a circle with center at point $O$ be given. Two different tangents are drawn from point $A$ (which lies on the entire circle). From the point of contact $B$ and $C$, respectively (Fig. 3).

Let us prove that $\angle BAO=\angle CAO$ and that $AB=AC$.

Figure 3. Illustration of Theorem 3

By Theorem 1, we have:

Therefore, the triangles $ABO$ and $ACO$ are right triangles. Since $OB=OC=r$, and the hypotenuse $OA$ is common, then these triangles are equal in hypotenuse and leg.

Hence we get that $\angle BAO=\angle CAO$ and $AB=AC$.

The theorem has been proven.

Example of a problem on the concept of a tangent to a circle

Example 1

Given a circle with center at point $O$ and radius $r=3\ cm$. The tangent $AC$ has a point of tangency $C$. $AO=4\ cm$. Find $AC$.

Solution.

Let's first depict everything in the figure (Fig. 4).

Figure 4.

Since $AC$ is a tangent and $OC$ is a radius, then by Theorem 1, we obtain that $\angle ACO=(90)^(()^\circ )$. We found that the triangle $ACO$ is rectangular, which means, by the Pythagorean theorem, we have:

\[(AC)^2=(AO)^2+r^2\] \[(AC)^2=16+9\] \[(AC)^2=25\] \

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, carried to the bases 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 equals 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.

Segments of tangents 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. PROOF. A. 3. B. 4. 1. 2. S. O. By the theorem about the tangent property, angles 1 and 2 are right angles, therefore triangles ABO and ACO are right-angled. They are equal, because have a common hypotenuse OA and equal legs OV and OS. Therefore, AB = AC and angle 3 = angle 4, which is what needed to be proven.

Slide 4 from the presentation "Circle" geometry.

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A straight line relative to a circle can be in the following three positions:

1. The distance from the center of the circle to the straight line is greater than the radius. In this case, all points of the line lie outside the circle.


2. The distance from the center of the circle to the straight line is less than the radius. In this case, the straight line has points lying inside the circle and since the straight line is infinite in both directions, it is intersected by the circle at 2 points.


3. The distance from the center of the circle to the straight line is equal to the radius. Straight line is tangent.

A straight line that has only one point in common with a circle is called tangent to the circle.

The common point is called in this case point of contact.

The possibility of the existence of a tangent, and, moreover, drawn through any point of the circle as a point of tangency, is proved by the following theorem.

Theorem. If a line is perpendicular to the radius at its end lying on the circle, then this line is a tangent.

Let O (fig) be the center of some circle and OA some of its radius. Through its end A we draw MN ^ OA.

It is required to prove that the line MN is tangent, i.e. that this line has only one common point A with the circle.

Let us assume the opposite: let MN have another common point with the circle, for example B.

Then straight line OB would be a radius and therefore equal to OA.

But this cannot be, since if OA is perpendicular, then OB must be inclined to MN, and the inclined one is greater than the perpendicular.

Converse theorem. If a line is tangent to a circle, then the radius drawn to the point of tangency is perpendicular to it.

Let MN be the tangent to the circle, A the point of tangency, and O the center of the circle.

It is required to prove that OA^MN.

Let's assume the opposite, i.e. Let us assume that the perpendicular dropped from O to MN will not be OA, but some other line, for example, OB.

Let's take BC = AB and carry out OS.

Then OA and OS will be inclined, equally distant from the perpendicular OB, and therefore OS = OA.

It follows from this that the circle, taking into account our assumption, will have two common points with the line MN: A and C, i.e. MN will not be a tangent, but a secant, which contradicts the condition.

Consequence. Through any given point on a circle one can draw a tangent to this circle, and only one, since through this point one can draw a perpendicular, and only one, to the radius drawn into it.

Theorem. A tangent parallel to a chord divides the arc subtended by the chord in half at the point of contact.

Let straight line AB (fig.) touch the circle at point M and be parallel to chord CD.

We need to prove that ÈCM = ÈMD.

Drawing the diameter ME through the point of tangency, we obtain: EM ^ AB, and therefore EM ^ CB.

Therefore CM=MD.

Task. Through a given point draw a tangent to a given circle.

If given point is on a circle, then draw a radius through it and a perpendicular straight line through the end of the radius. This line will be the desired tangent.

Let us consider the case when the point is given outside the circle.

Let it be required (Fig.) to draw a tangent to a circle with center O through point A.

To do this, from point A, as the center, we describe an arc with radius AO, and from point O, as the center, we intersect this arc at points B and C with a compass opening equal to the diameter of the given circle.

Having then drawn the chords OB and OS, we connect point A with points D and E, at which these chords intersect with the given circle.

Lines AD and AE are tangent to circle O.

Indeed, from the construction it is clear that the pipes AOB and AOC are isosceles (AO = AB = AC) with the bases OB and OS equal to the diameter of the circle O.

Since OD and OE are radii, then D is the middle of OB, and E is the middle of OS, which means AD and AE are medians drawn to the bases of isosceles pipes, and therefore perpendicular to these bases. If the lines DA and EA are perpendicular to the radii OD and OE, then they are tangent.

Consequence. Two tangents drawn from one point to a circle are equal and form equal angles with the straight line connecting this point to the center.

So AD=AE and ÐOAD = ÐOAE (Fig.), because rectangular tr-ki AOD and AOE, having a common hypotenuse AO and equal legs OD and OE (as radii), are equal.

Note that here the word “tangent” means the actual “tangent segment” from a given point to the point of contact.

Task. Draw a tangent to a given circle O parallel to a given straight line AB (Fig.).

We lower a perpendicular OS to AB from the center O and through the point D, at which this perpendicular intersects the circle, draw EF || AB.

The tangent we are looking for will be EF.

Indeed, since OS ^ AB and EF || AB, then EF ^ OD, and the line perpendicular to the radius at its end lying on the circle is a tangent.

Task. Draw a common tangent to two circles O and O 1 (Fig.).

Analysis. Let's assume that the problem is solved.

Let AB be the common tangent, A and B the points of tangency.

Obviously, if we find one of these points, for example, A, then we can easily find the other one.

Let us draw the radii OA and O 1 B. These radii, being perpendicular to the common tangent, are parallel to each other.

Therefore, if from O 1 we draw O 1 C || BA, then the pipeline OCO 1 will be rectangular at vertex C.

As a result, if we describe a circle from O as the center with radius OS, then it will touch the straight line O 1 C at point C.

The radius of this auxiliary circle is known: it is equal to OA – CA = OA - O 1 B, i.e. it is equal to the difference between the radii of these circles.

Construction. From the center O we describe a circle with a radius equal to the difference given radii.

From O 1 we draw a tangent O 1 C to this circle (in the manner indicated in the previous problem).

Through the tangent point C we draw the radius OS and continue it until it meets the given circle at point A. Finally, from A we draw AB parallel to CO 1.

In exactly the same way we can construct another common tangent A 1 B 1 (Fig.). Direct lines AB and A 1 B 1 are called external common tangents.

You can spend two more internal tangents as follows:

Analysis. Let's assume that the problem is solved (Fig.). Let AB be the desired tangent.

Let us draw the radii OA and O 1 B to the tangent points A and B. Since these radii are both perpendicular to the common tangent, they are parallel to each other.

Therefore, if from O 1 we draw O 1 C || BA and continue OA to point C, then OS will be perpendicular to O 1 C.

As a result, the circle described by the radius OS from point O as the center will touch the straight line O 1 C at point C.

The radius of this auxiliary circle is known: it is equal to OA+AC = OA+O 1 B, i.e. it is equal to the sum of the radii of the given circles.

Construction. From O as the center, we describe a circle with a radius equal to the amount given radii.

From O 1 we draw a tangent O 1 C to this circle.

We connect the point of contact C with O.

Finally, through point A, at which OS intersects the given circle, we draw AB = O 1 C.

In a similar way we can construct another internal tangent A 1 B 1.

General definition of tangent

Let a tangent AT and some secant AM be drawn through point A to a circle with a center (Fig.).

Let's rotate this secant around point A so that the other intersection point B moves closer and closer to A.

Then the perpendicular OD, lowered from the center to the secant, will approach the radius OA more and more, and the angle AOD may become less than any small angle.

The angle MAT formed by a secant and a tangent is equal to angle AOD (due to the perpendicularity of their sides).

Therefore, as point B approaches A indefinitely, angle MAT can also become arbitrarily small.

This is expressed in other words like this:

a tangent is the limiting position to which a secant drawn through a point of tangency tends when the second point of intersection approaches the point of tangency indefinitely.

This property is taken to be the definition of a tangent when we're talking about about any curve.

Thus, the tangent to the curve AB (Fig.) is the limiting position MT to which the secant MN tends when the intersection point P approaches M without limit.

Note that the tangent defined in this way can have more than one common point with the curve (as can be seen in Fig.).