Junior high school math problem about the relationship between the position of a straight line and a

Proof: Connect the OC

Because oa=oc=ob

So aco= bac=30°

and ab is the diameter of the circle, so acb=90°

ME vertical ab

So emb=90°

So ecf= bac=30°

ecf= e

So ecf=30°

then fcn=90-30=60°

So fco= fcn+ aco=90°

i.e. CF Vertical OC

So cf is the tangent of the circle o.

2. If the radius of the circle o is 1, then ab=2

ac=√3bc=1

So ce= 3

mo=be*sine-ob=1/2(1+√3)-1=1/2(√3-1)

ab is the circle diameter, so there is ac be, oc is abc midline, oca = a = 30°

em⊥ab，∠e=∠a

ecf=∠e=30°

So cf oc, c is on the circle, and cf is the tangent of the circle.

By the abc ecn emb

ab=2r=2

ac=√3bc=ob=oc=1

bm:bc=ab:be

be=bc+ce=bc+ac

bm:1=2:(1+√3)

bm=2/(1+√3)

om=bm-1=2/(1+√3)-1

Question 1: OA=5 3

Question 2: d, the distance to the center of the circle is equal to the radius of the straight line is the tangent of the circle Question 3: c, r

Question 4: Proof: Nexus be, because ab is the diameter.

So be vertical ac

In the RT triangle AEB and the RT triangle BEC.

O and D are the midpoints of the hypotenuse AB and BC, respectively.

So oe=ob

db=de, so again

2.(1) Connect AO because the angle b = 30 degrees.

So the angle AOC = 60 degrees. And because AO=CO, the triangle AOC is an equilateral triangle, i.e., the angle OCA=60 degrees, the angle ACD=120 degrees, and because the angle CAD=30 degrees, the angle D=30 degrees.

Angle AOC = 60 degrees and angle D = 30 degrees.

So the angle oad = 90 degrees.

i.e. ao ad

So ad is the tangent of the circle o.

Proven wrong.

The radius is r

So the perimeter a=2pi*r

Area a=pi*r*r

So 2r=r*r, r=2 is the distance x from the straight line to the center of the circle.

When x=r, tangent.

XR is separated. This can be seen from the figure.

The relationship between a straight line and a circle: apart, tangent, intersecting.

Separation: d>r

Tangent: d=r

Intersection: d

Proof: Connect the OC

Because oa=oc=ob

So aco= bac=30°

and ab is the diameter of the circle, so acb=90°

me vertical ab so emb=90°

So ecf= bac=30°

ecf= e

So ecf=30°

then fcn=90-30=60°

So fco= fcn+ aco=90°

i.e. CF is perpendicular to OC, so CF is the tangent of the circle O.

2. If the radius of the circle o is 1, then ab=2

ac= 3 bc=1 so ce= 3

mo=be*sine-ob=1/2(1+√3)-1=1/2(√3-1)

The relationship between the position of the straight line and the circle in the second semester of the third semester of junior high school (2) Practice question 1: Basic training 1 A tangent line can be made from a point on the circle; A little outside the circle can be made as a tangent line of the circle; The tangent of the circle that is buried a little bit in the circle is slowed down

2 If one side of the triangle is straight, which circle of the diameter of the ant is exactly tangent to the other side, then the triangle is 3 The following straight line is the tangent of the circle, and the tangent is ( ).

A A line with a common point with a circle b The distance from the center of the circle is equal to the radius of the line c The line perpendicular to the radius of the circle d The line at the outer end of the diameter of the circle 4 oa bisects boc, p is any point on oa (except o), if p with p as the center of the circle is tangent to oc, then the position position of p and ob is ( ).

A intersects B tangent c distancs d intersects or tangents.

1) The diameter is 12, then the radius r is 6 r>d, and the line intersects the circle with two intersections.

2) As od ab, then sin oab=od oa od= 3o, the distance from ab is the same as the radius of the circle.

The circle is tangent to the straight line ab.

There are two intersections, according to the definition of a straight line and a circle are tangent, the second question is a bit incomprehensible, if my understanding is like this, it should be separated.

There are two common points where a line intersects a circle.

Solution: When the chord ab is bisected by the point p, obviously.

AB is bisected perpendicularly by the diameter op.

The analytic formula for finding the straight line op is: y

2x So, the analytic formula for the straight line ab is: y

2(x+1)

Asking for the length of the string is trivial, I don't think it needs to be said!

In this figure, the radius of the inscribed circle = 1 3 equilateral triangles are high.

And equilateral triangle height = 2 * cos30° = 3

So the radius is 3 3 long (just read it wrong, sorry).

The side length is 2, then the height is the root number 3, and OA is equal to 2 times OD, so OD is equal to one-third of the root number 3, and the radius is one-third of the root number 3

According to the triangle odc is a right triangle, the angle ocd is 30°, so 0d is equal to cd root number 3, cd = 1, so od = root number 3 3