1. As shown in the figure, AB is the diameter ⊙O, CD is the tangent ⊙O, and the tangent point is D. The extension lines of CD and AB intersect at point C. A=30？ The following three conclusions are given: ① AD = CD; ②BD = BC； ③AB=2BC, in which the number of correct conclusions is ()

A. 1 D. 0

Test center: the nature of tangent.

Analysis: connecting OD, CD is tangent to ⊙O, can you get CD? OD，by？ A=30？ Can you draw? ABD=60？ △ODB is an equilateral triangle. C=？ BDC=30？ Then in a right triangle, the right side opposite to 300 is equal to half of the hypotenuse, so the conclusion ① ② ③ holds.

Solution: solution: as shown in the figure, connect OD,

∵CD is the tangent of⊙ O,

? CD？ OD，

ODC=90？ ,

Again? A=30？ ,

ABD=60？ ,

? △OBD is an equilateral triangle,

DOB=？ ABD=60？ ，AB = 2OB = 2OD = 2BD。

C=？ BDC=30？ ,

? BD=BC, ② holds;

? AB=2BC, ③ holds;

A=？ c，

? DA=DC, ① holding;

To sum up, ① ② ③ all hold,

So, the answer is a.

Comments: This topic examines the comprehensive application of the related properties of the circle. It is the key to solve the problem to find the degree of the corresponding angle by borrowing the nature of the tangent.

2. As shown in the figure, the length of the rectangular ABCD is 6, the width is 3, the point O 1 is the center of the rectangle, and the radius ⊙ O2 is 1, O 1O2? AB, o1O2 at point p = 6. If ⊙O2 rotates 360 clockwise around point P? During the rotation, there is only one common point between ⊙O2 and the edge of the rectangle, and a * * * sign () appears.

A.3 times B. 4 times C. 5 times D. 6 times

Test center: the positional relationship between straight line and circle.

Analysis: Make a picture according to the meaning of the question and write the answer directly.

Solution: Solution: As shown in the figure ⊙O2 and the sides of the rectangle have only one thing in common. A * * * appears four times.

So choose B.

Comments: This question examines the positional relationship between a straight line and a circle. The key to solving the problem is to understand that when a circle is tangent to a straight line, the distance from the point to the center of the circle is equal to the radius of the circle.

3. As shown in the figure, in the plane rectangular coordinate system xOy, the coordinate of the center p of the radius ⊙P is (3,0). If ⊙P translates along the positive direction of the X axis so that ⊙P is tangent to the Y axis, the translation distance is ().

(DrawingNo. 1)

A. 1 B. 1 or 5c.3d.5

Test center: the positional relationship between straight line and circle; Coordinates and graphic properties.

Analysis: translation is divided into two cases: the left side of Y axis and the right side of Y axis.

Solution: Solution: When ⊙P is located on the left side of the Y axis and tangent to the Y axis, the translation distance is1;

When ⊙P is located on the right side of the Y axis and tangent to the Y axis, the translation distance is 5.

So choose B.

Comments: This question examines the positional relationship between a straight line and a circle. The key to solving the problem is to understand that when a circle is tangent to a straight line, the distance from the point to the center of the circle is equal to the radius of the circle.

4. As shown in the figure, P is a point on the extension line of BA, the diameter is ⊙O, PC is tangent to ⊙O, the tangent point is C, and the point D is the upper point of ⊙, connecting PD. Known PC=PD=BC. Draw the following conclusions:

(1)PD is tangent to ⊙O; (2) The quadrangle PCBD is a diamond; (3)PO = AB； (4)? PDB= 120？ .

The correct number is ()

A.4b 3c 2d 1

Analysis: (1) Using the properties of tangents? PCO=90？ , and then get △ PCO △ PDO (SSS), which can be obtained? PCO=？ PDO=90？ Get the answer;

(2) Use (1) to get:? CPB=？ BPD, and then find △ CPB △ DPB (SAS), you can get the answer;

(3) △ PCO △ BCA (ASA) is obtained through congruent triangles's judgment, and then CO = PO = AB is obtained.

(4) The quadrangle PCBD is a diamond. CPO=30？ , then DP=DB, then? DPB=？ DBP=30？ , you will know.

Solution: (1) Connect co, DO,

∵PC is tangent to⊙ O, and the tangent point is C, PCO=90? ,

In △PCO and △PDO, △ PCO △ PDO (SSS), PCO=? PDO=90？ ,

? PD is tangent to ⊙O, so the option is correct;

(2) From ( 1):？ CPB=？ BPD，

In delta CPB and delta DPB, delta CPB delta DPB (SAS),

? BC=BD，？ PC=PD=BC=BD，？ The quadrangle PCBD is a diamond, so the option is correct;

(3) connect communication,

PC = CB，CPB=？ CBP, ∫AB is the diameter ⊙O, ACB=90? ,

In △PCO and △BCA, △ PCO △ BCA (ASA),

? AC=CO，？ AC=CO=AO，COA=60？ ，CPO=30？ ,

? CO= PO= AB，？ PO=AB, so the option is correct;

(4) The quadrangle PCBD is a diamond. CPO=30？ ,

? DP=DB, then? DPB=？ DBP=30？ ，PDB= 120？ , so the option is correct; So choose: a.

Comments: This question mainly examines the judgment and nature of tangent, congruent triangles and diamond, and skillfully uses congruent triangles's judgment and nature to solve problems.

5.(20 14? Wuhan, question 10, 3 points) As shown in the figure, PA and PB cut ⊙O at point A and point B, CD cut ⊙O at point E, and PA and PB cross at point C and point D. If the radius of ⊙O is r and the circumference of △PCD is 3r, then tan? The value of APB is ()

A. 1

1/2

C.3/5

D.2

Test center: the nature of tangent; Similar triangles's judgment and nature; Definition of acute angle trigonometric function

Analysis: (1) connect OA, OB and OP, extend the extension line of BO to PA at point F, and get CA=CE, DB=DE, PA=PB by tangent, and then get PA = Pb. Use Rt△BFP∽RT△OAF to get AF= FB. In RT△FBP, use Pythagorean theorem to find BF. The value of APB is enough.

Solution: connect OA, OB and OP, and extend the extension line of BO to PA at point F.

Pa, PB cut ≧O at point A and point B, and CD cut ≧O at point E.

OAP=？ OBP=90？ ，CA=CE，DB=DE，PA=PB，

∵△ Perimeter ∵△ PCD = PC+CE+DE+PD = PC+AC+PD+DB = PA+Pb = 3r,

? PA=PB=。

In Rt△BFP and Rt△OAF,

,

? Rt△BFP∽RT△OAF。

? = = = ,

? AF= FB，

At Rt△FBP,

∵PF2﹣PB2=FB2

? (PA+AF)2﹣PB2=FB2

? (r+ BF)2﹣( )2=BF2，

The solution is BF= r,

? Tan? APB= = =，

Therefore, choose: B.

6. As shown in the figure, G is the center of gravity of △ABC. If the circle G is tangent to AC and BC respectively, and intersects with AB at two points, which of the following is true about the sizes of the three sides of △ABC? ( )

A.BCAC·ABAC

Analysis: G is the center of gravity of △ABC, so △ABG area =△BCG area =△ACG area can be judged according to the triangle area formula.

Solution: ∫G is the center of gravity of △ABC,

? △ABG area =△ BCG area =△ACG area,

GHa = GHb & gtGHc，

? BC=AC

So choose D.

Comments: This question examines the nature of the center of gravity of the triangle and the area formula of the triangle. Understanding the nature of the center of gravity is the key.

7. As shown in the figure, in ⊙O with a radius of 6cm, point A is the midpoint of the lower arc, point D is the point on the upper arc, and? D=30？ , draw the following four conclusions:

①OA？ BC; ②BC = 6； 3 sin? AOB =； ④ The quadrilateral ABOC is a diamond.

The serial number of the correct conclusion is ()

A.①③ B. ①②③④ C. ②③④ D. ①③④

Test site: vertical diameter theorem; Determination of diamond shape; Theorem of circle angle; Solve a right triangle.

Analysis: According to the definition of vertical diameter theorem, diamond judgment theorem and acute angle trigonometric function, each option can be judged one by one.

Solution: Solution: ∵ Point A is the midpoint of the bad arc and OA passes through the center of the circle.

? OA？ BC, so ① is correct;

∵? D=30？ ,

ABC=？ D=30？ ,

AOB=60？ ,

Point a is the midpoint of the bad arc,

? BC=2CE，

OA = OB，

? OB=OB=AB=6cm，

? BE=AB？ cos30？ =6? =3 cm,

? BC=2BE=6 cm, so B is correct;

∵? AOB=60？ ,

? Sin? AOB=sin60？ = ,

Therefore, 3 is correct;

∵? AOB=60？ ,

? AB=OB，

Point a is the midpoint of the bad arc,

? AC=OC，

? AB=BO=OC=CA，

? The quadrilateral ABOC is a diamond,

Therefore, 4 is correct.

So choose B.

Comments: This topic examines the vertical diameter theorem, the judgment of diamond, the fillet theorem and the solution of right triangle, which is comprehensive and good.

8. As shown in the figure, in the plane rectangular coordinate system, the central coordinate ⊙P is (3, a) (a >; 3), the radius is 3, and the length of the chord AB of the image with the function y=x divided by ⊙P is, then the value of a is ().

A.4 B. 7C.3 D.5

Answer: Solution: Do PC? X axis in c, cross AB and PE in d? AB in E is connected with PB, as shown in the figure,

The central coordinate of p is (3, a),

? OC=3，PC=a，

Substitute x=3 into y=x to get y=3.

? The coordinates of point d are (3,3),

? CD=3，

? △OCD is an isosceles right triangle,

? △PED is also an isosceles right triangle

∵PE？ AB，

? AE=BE=AB=？ 4 =2 ,

In Rt△PBE, PB=3,

? PE=，

? PD= PE=，

? a=3+。

So choose B.

Comments: This question examines the vertical diameter theorem: bisect the diameter of a chord, bisect this chord, and bisect the two arcs opposite to this chord. It also examines the Pythagorean theorem and the properties of isosceles right triangle.