Let us recall an important definition - the definition of a circle]
Definition:
A circle with center at point O and radius R is the set of all points of the plane located at a distance R from point O.
Let us pay attention to the fact that a circle is a set everyone points satisfying the described condition. Let's look at an example:
Points A, B, C, D of the square are equidistant from point E, but they are not a circle (Fig. 1).
Rice. 1. Illustration for example
In this case, the figure is a circle, since it is all a set of points equidistant from the center.
If you connect any two points on a circle, you get a chord. The chord passing through the center is called the diameter.
MB - chord; AB - diameter; MnB is an arc, it is contracted by the MV chord;
The angle is called central.
Point O is the center of the circle.
Rice. 2. Illustration for example
Thus, we remembered what a circle is and its main elements. Now let's move on to considering the relative position of the circle and the straight line.
Given a circle with center O and radius r. Straight line P, the distance from the center to the straight line, that is, perpendicular to OM, is equal to d.
We assume that point O does not lie on line P.
Given a circle and a straight line, we need to find the number of common points.
Case 1 - the distance from the center of the circle to the straight line is less than the radius of the circle:
In the first case, when the distance d is less than the radius of the circle r, point M lies inside the circle. From this point we will plot two segments - MA and MB, the length of which will be . We know the values of r and d, d is less than r, which means the expression exists and points A and B exist. These two points lie on a straight line by construction. Let's check if they lie on the circle. Let us calculate the distance OA and OB using the Pythagorean theorem:
Rice. 3. Illustration for case 1
The distance from the center to two points is equal to the radius of the circle, so we have proven that points A and B belong to the circle.
So, points A and B belong to the line by construction, they belong to the circle by what has been proven - the circle and the line have two common points. Let us prove that there are no other points (Fig. 4).
Rice. 4. Illustration for the proof
To do this, take an arbitrary point C on a straight line and assume that it lies on a circle - distance OS = r. In this case, the triangle is isosceles and its median ON, which does not coincide with the segment OM, is the height. We get a contradiction: two perpendiculars are dropped from point O onto a straight line.
Thus, there are no other common points on the line P with the circle. We have proven that in the case where the distance d is less than the radius of the circle r, the straight line and the circle have only two points in common.
Case two - the distance from the center of the circle to the straight line is equal to the radius of the circle (Fig. 5):
Rice. 5. Illustration for case 2
Recall that the distance from a point to a straight line is the length of the perpendicular, in this case OH is the perpendicular. Since, by condition, the length OH is equal to the radius of the circle, then point H belongs to the circle, thus point H is common to the line and the circle.
Let us prove that there are no other common points. By contrast: suppose that point C on the line belongs to the circle. In this case, the distance OS is equal to r, and then OS is equal to OH. But in a right triangle, the hypotenuse OC is greater than the leg OH. We got a contradiction. Thus, the assumption is false and there is no point other than H that is common to the line and the circle. We have proven that in this case there is only one common point.
Case 3 - the distance from the center of the circle to the straight line is greater than the radius of the circle:
The distance from a point to a line is the length of the perpendicular. We draw a perpendicular from point O to line P, we get point H, which does not lie on the circle, since OH is by condition greater than the radius of the circle. Let us prove that any other point on the line does not lie on the circle. This is clearly seen from a right triangle, the hypotenuse OM of which is greater than the leg OH, and therefore greater than the radius of the circle, thus point M does not belong to the circle, like any other point on the line. We have proven that in this case the circle and the straight line do not have common points (Fig. 6).
Rice. 6. Illustration for case 3
Let's consider theorem . Let us assume that straight line AB has two common points with the circle (Fig. 7).
Rice. 7. Illustration for the theorem
We have a chord AB. Point H, by convention, is the middle of the chord AB and lies on the diameter CD.
It is required to prove that in this case the diameter is perpendicular to the chord.
Proof:
Consider the isosceles triangle OAB, it is isosceles because .
Point H, by convention, is the midpoint of the chord, which means the midpoint of the median AB of an isosceles triangle. We know that the median of an isosceles triangle is perpendicular to its base, which means it is the height: , hence, thus, it is proven that the diameter passing through the middle of the chord is perpendicular to it.
Fair and converse theorem : if the diameter is perpendicular to the chord, then it passes through its middle.
Given a circle with center O, its diameter CD and chord AB. It is known that the diameter is perpendicular to the chord; it is necessary to prove that it passes through its middle (Fig. 8).
Rice. 8. Illustration for the theorem
Proof:
Consider the isosceles triangle OAB, it is isosceles because . OH, by convention, is the height of the triangle, since the diameter is perpendicular to the chord. The height in an isosceles triangle is also the median, so AN = HB, which means that point H is the midpoint of the chord AB, which means that it is proven that the diameter perpendicular to the chord passes through its midpoint.
The direct and converse theorem can be generalized as follows.
Theorem:
A diameter is perpendicular to a chord if and only if it passes through its midpoint.
So, we have considered all cases of the relative position of a line and a circle. In the next lesson we will look at the tangent to a circle.
Bibliography
- Alexandrov A.D. etc. Geometry 8th grade. - M.: Education, 2006.
- Butuzov V.F., Kadomtsev S.B., Prasolov V.V. Geometry 8. - M.: Education, 2011.
- Merzlyak A.G., Polonsky V.B., Yakir S.M. Geometry 8th grade. - M.: VENTANA-GRAF, 2009.
- Edu.glavsprav.ru ().
- Webmath.exponenta.ru ().
- Fmclass.ru ().
Homework
Task 1. Find the lengths of two segments of the chord into which the diameter of the circle divides it, if the length of the chord is 16 cm and the diameter is perpendicular to it.
Task 2. Indicate the number of common points of a line and a circle if:
a) the distance from the straight line to the center of the circle is 6 cm, and the radius of the circle is 6.05 cm;
b) the distance from the straight line to the center of the circle is 6.05 cm, and the radius of the circle is 6 cm;
c) the distance from the straight line to the center of the circle is 8 cm, and the radius of the circle is 16 cm.
Task 3. Find the length of the chord if the diameter is perpendicular to it, and one of the segments cut off by the diameter from it is 2 cm.
Let us recall an important definition - the definition of a circle]
Definition:
A circle with center at point O and radius R is the set of all points of the plane located at a distance R from point O.
Let us pay attention to the fact that a circle is a set everyone points satisfying the described condition. Let's look at an example:
Points A, B, C, D of the square are equidistant from point E, but they are not a circle (Fig. 1).
Rice. 1. Illustration for example
In this case, the figure is a circle, since it is all a set of points equidistant from the center.
If you connect any two points on a circle, you get a chord. The chord passing through the center is called the diameter.
MB - chord; AB - diameter; MnB is an arc, it is contracted by the MV chord;
The angle is called central.
Point O is the center of the circle.
Rice. 2. Illustration for example
Thus, we remembered what a circle is and its main elements. Now let's move on to considering the relative position of the circle and the straight line.
Given a circle with center O and radius r. Straight line P, the distance from the center to the straight line, that is, perpendicular to OM, is equal to d.
We assume that point O does not lie on line P.
Given a circle and a straight line, we need to find the number of common points.
Case 1 - the distance from the center of the circle to the straight line is less than the radius of the circle:
In the first case, when the distance d is less than the radius of the circle r, point M lies inside the circle. From this point we will plot two segments - MA and MB, the length of which will be . We know the values of r and d, d is less than r, which means the expression exists and points A and B exist. These two points lie on a straight line by construction. Let's check if they lie on the circle. Let us calculate the distance OA and OB using the Pythagorean theorem:
Rice. 3. Illustration for case 1
The distance from the center to two points is equal to the radius of the circle, so we have proven that points A and B belong to the circle.
So, points A and B belong to the line by construction, they belong to the circle by what has been proven - the circle and the line have two common points. Let us prove that there are no other points (Fig. 4).
Rice. 4. Illustration for the proof
To do this, take an arbitrary point C on a straight line and assume that it lies on a circle - distance OS = r. In this case, the triangle is isosceles and its median ON, which does not coincide with the segment OM, is the height. We get a contradiction: two perpendiculars are dropped from point O onto a straight line.
Thus, there are no other common points on the line P with the circle. We have proven that in the case where the distance d is less than the radius of the circle r, the straight line and the circle have only two points in common.
Case two - the distance from the center of the circle to the straight line is equal to the radius of the circle (Fig. 5):
Rice. 5. Illustration for case 2
Recall that the distance from a point to a straight line is the length of the perpendicular, in this case OH is the perpendicular. Since, by condition, the length OH is equal to the radius of the circle, then point H belongs to the circle, thus point H is common to the line and the circle.
Let us prove that there are no other common points. By contrast: suppose that point C on the line belongs to the circle. In this case, the distance OS is equal to r, and then OS is equal to OH. But in a right triangle, the hypotenuse OC is greater than the leg OH. We got a contradiction. Thus, the assumption is false and there is no point other than H that is common to the line and the circle. We have proven that in this case there is only one common point.
Case 3 - the distance from the center of the circle to the straight line is greater than the radius of the circle:
The distance from a point to a line is the length of the perpendicular. We draw a perpendicular from point O to line P, we get point H, which does not lie on the circle, since OH is by condition greater than the radius of the circle. Let us prove that any other point on the line does not lie on the circle. This is clearly seen from a right triangle, the hypotenuse OM of which is greater than the leg OH, and therefore greater than the radius of the circle, thus point M does not belong to the circle, like any other point on the line. We have proven that in this case the circle and the straight line do not have common points (Fig. 6).
Rice. 6. Illustration for case 3
Let's consider theorem . Let us assume that straight line AB has two common points with the circle (Fig. 7).
Rice. 7. Illustration for the theorem
We have a chord AB. Point H, by convention, is the middle of the chord AB and lies on the diameter CD.
It is required to prove that in this case the diameter is perpendicular to the chord.
Proof:
Consider the isosceles triangle OAB, it is isosceles because .
Point H, by convention, is the midpoint of the chord, which means the midpoint of the median AB of an isosceles triangle. We know that the median of an isosceles triangle is perpendicular to its base, which means it is the height: , hence, thus, it is proven that the diameter passing through the middle of the chord is perpendicular to it.
Fair and converse theorem : if the diameter is perpendicular to the chord, then it passes through its middle.
Given a circle with center O, its diameter CD and chord AB. It is known that the diameter is perpendicular to the chord; it is necessary to prove that it passes through its middle (Fig. 8).
Rice. 8. Illustration for the theorem
Proof:
Consider the isosceles triangle OAB, it is isosceles because . OH, by convention, is the height of the triangle, since the diameter is perpendicular to the chord. The height in an isosceles triangle is also the median, so AN = HB, which means that point H is the midpoint of the chord AB, which means that it is proven that the diameter perpendicular to the chord passes through its midpoint.
The direct and converse theorem can be generalized as follows.
Theorem:
A diameter is perpendicular to a chord if and only if it passes through its midpoint.
So, we have considered all cases of the relative position of a line and a circle. In the next lesson we will look at the tangent to a circle.
Bibliography
- Alexandrov A.D. etc. Geometry 8th grade. - M.: Education, 2006.
- Butuzov V.F., Kadomtsev S.B., Prasolov V.V. Geometry 8. - M.: Education, 2011.
- Merzlyak A.G., Polonsky V.B., Yakir S.M. Geometry 8th grade. - M.: VENTANA-GRAF, 2009.
- Edu.glavsprav.ru ().
- Webmath.exponenta.ru ().
- Fmclass.ru ().
Homework
Task 1. Find the lengths of two segments of the chord into which the diameter of the circle divides it, if the length of the chord is 16 cm and the diameter is perpendicular to it.
Task 2. Indicate the number of common points of a line and a circle if:
a) the distance from the straight line to the center of the circle is 6 cm, and the radius of the circle is 6.05 cm;
b) the distance from the straight line to the center of the circle is 6.05 cm, and the radius of the circle is 6 cm;
c) the distance from the straight line to the center of the circle is 8 cm, and the radius of the circle is 16 cm.
Task 3. Find the length of the chord if the diameter is perpendicular to it, and one of the segments cut off by the diameter from it is 2 cm.
Let's take an arbitrary circle with a center at point O and a straight line a.
If straight line a passes through point O, then it will intersect the given circle at two points K and L, which are the ends of the diameter lying on straight line a.
If straight line a does not pass through the center O of the circle, then we will perform an auxiliary construction and draw a straight line OH perpendicular to a straight line a and denote the resulting distance from the center of the circle to the straight line a variable rasstoyanie. Let's determine how many common points the line will have a and circles depending on the relationship between the variable rasstoyanie and radius.
There may be 3 options:
- rasstoyanie < radius. In this case, the point H will lie in the middle of the circle, which is limited by the given circle.
Let's put a segment on a straight line HD = radius.
In OHD the hypotenuse O.D. more leg HD, That's why OD > radius. Therefore, the point D lies beyond the circle bounded by the given circle. So one end of the segment HD is in the middle of the circle, and the other is outside the circle. Thus, on the segment HD you can mark a point A, which lies on the circle, that is OA = radius.
Let's extend the beam H.A. and put a segment on it BH, which is equal to the segment AN.
Received 2 right triangles OHA And OHB, which are equal on two legs. Then their corresponding sides are equal: OB = OA = r. Hence, B is also the common point of a circle and a line. Since 3 points of a circle cannot lie on the same line, then other common points of the line a and circles do not exist.
Thus, if the distance between the center of the circle and the straight line is less than the radius of the circle ( rasstoyanie < r
adius), then the line and the circle have 2 common points.
- rasstoyanie= radius . Because the OH = radius, then point H belongs to the circle and is therefore a common point for the line a and circles.
For any other points on the line a(for example, points and M) oblique OM more segment OH, that is OM > OH = radius, and therefore the point M does not belong to the given circle.
Therefore, if the distance between the center of the circle and the straight line is equal to the radius of the circle ( rasstoyanie= radius), then the line and the circle have only one common point.
- rasstoyanie>radius . Since OH > radius, then for any points of the line a(for example, points M) the inequality holds OM > OH > radius. So the point M does not belong to the circle.
Therefore, if the distance between the center of the circle and the straight line is greater than the radius of the circle ( rasstoyanie>radius), then the line and the circle have no common points.
Compiled by a math teacher
MBOU Secondary School No. 18, Krasnoyarsk
Andreeva Inga Viktorovna
The relative position of a straight line and a circle
ABOUT R – radius
WITH D – diameter
AB- chord
- Circle with center at a point ABOUT radius r
- A straight line that does not pass through the center ABOUT
- Let us denote the distance from the center of the circle to the straight line by the letter s
Three cases are possible:
- 1) s
- less radius of the circle, then the straight line and the circle have two common points .
Direct AB is called secant in relation to the circle.
Three cases are possible:
- 2 ) s = r
- If the distance from the center of the circle to the straight line equals radius of the circle, then the straight line and the circle have only one common point .
s = r
r 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 do not have common points. sr r O" width="640" Three cases are possible:
- 3 ) sr
- If the distance from the center of the circle to the straight line more radius of a circle, then a straight line and a circle have no common points .
Tangent to a circle
Definition: P a 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.
s = r
- straight line - secant
- straight line - secant
- no common points
- straight line - secant
- straight line - tangent
- r = 15 cm, s = 11 cm
- r = 6 cm, s = 5.2 cm
- r = 3.2 m, s = 4.7 m
- r = 7 cm, s = 0.5 dm
- r = 4 cm, s = 4 0 mm
Solve No. 633.
- OABC-square
- AB = 6 cm
- Circle with center O of radius 5 cm
secants from straight lines OA, AB, BC, AC
Tangent property: A tangent to a circle is perpendicular to the radius drawn to the point of tangency.
m– tangent to a circle with center ABOUT
M– point of contact
OM- radius
Tangent sign: If a straight line passes through the end of a radius lying on a circle and is perpendicular to the radius, then it is a asative.
circle with center ABOUT
radius OM
m- a straight line that passes through a point M
m – tangent
Property of tangents passing through one point:
Tangent segments to
circles drawn
from the same point, are equal and
make equal angles
with a straight line passing through
this point and the center of the circle.
▼ By the tangent property
∆ AVO, ∆ ASO – rectangular
∆ ABO= ∆ ACO – along the hypotenuse and leg:
OA - general,