Solution: As shown in the figure, the connection OD, ∵CD is tangent to ⊙O, ∴CD⊥OD, ∴∠ODC = 90°, ∴∠. ∠ A = 30 ∴∠.
2. The length of the rectangular ABCD is 6, the width is 3, the point O 1 is the center of the rectangle, the radius ⊙O2 is 1, O 1O2⊥AB is at the point P, O 1O2=6. If ⊙O2 rotates 360 clockwise around point P, there is only one common point between ⊙O2 and the edge of the rectangle during the rotation, and a * * * sign () appears.
Solution: ⊙O2 has only one common point with the edge of the rectangle-* * * appears four times, and the positional relationship between the straight line and the circle is investigated. 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. 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 ().
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. 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.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 a point above ⊙, connecting PD. Known PC=PD=BC. The following conclusions: PD and ⊙O phase; The quadrangle PCBD is a diamond; PO = AB∠PDB= 120. The correct number is ().
Answer: The connecting lines CO, DO, ∵PC are tangent to ⊙O, and the tangent point is C, ∴∠pco = 90°, which is between △PCO and △PDO, ∴△PCO≌△PDO(SSS), ∴∠.
5.PA and PB cut ⊙O at points A and B, CD cut ⊙O at point E, and PA and PB intersect at points C and D. If the radius of ⊙O is r and the circumference of △PCD is 3r, the value of tan∠APB is ().
Answer: Connect OA, OB and OP, and extend the extension line of BO to pa∵pa at point F, PB cut at point A and point B ⊙O, CD cut at point E ⊙O, ∴∠ OAP = ∠ OBP = 90, CA=CE, DB=DE. ∴(pa+af)22=fb2； ∴ (R+BF) 2 () 2 = BF2, the solution is BF=r, ∴ Tan ∠ APB = =.
6.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? () G is the center of gravity of △ABC, so △ABG area =△BCG area =△ACG area can be judged according to the triangle area formula.
Answer: ∫g is the center of gravity of △ABC, ∴△ABG area =△BCG area =△ACG area, and gha = ghb & gtGHc,∴BC=AC.