(1) Don't miss the solution of geometry problems without graphs.

In recent years, some geometry problems have not been given directly. Due to the influence of thinking habits, I did not carefully consider the conditions provided by the questions and lacked an accurate understanding of mathematical facts. Often, only one common figure that meets the requirements is considered, which leads to missing solutions. This kind of topic focuses on students' mastery and application of basic knowledge, which is conducive to cultivating rigorous logical thinking ability. If you don't think carefully when solving problems and form a mindset, you will miss the solution. Therefore, consider the problem comprehensively, such as:

1. The positional relationship of tangency between two circles includes inscribed circle and circumscribed circle.

2. When two circles are inscribed, the radii r 1 and r2 of the two circles are unknown, and the center d = | r 1-r2 | should be considered.

3. When the radius of two intersecting circles is known and the chord length is known, there are two positional relationships between the centers of the two circles and the chord: (1) The centers of the two circles are on both sides of the chord; (2) The two centers are on the same side of the chord.

4. The side of a right triangle can be the right side or the hypotenuse.

Example 1. It is known that the two sides of a right triangle are 3 and 4 respectively, and the length of the third side is = _ _ _ _ _.

Analysis: Most students are used to the saying "hook three strands, four chords and five", which means that when the two right angles are 3 and 4 respectively, the length of the hypotenuse is 5. But the premise of this understanding is that 3 and 4 are right angles. But this question is not explained, so the third side may be a hypotenuse, but it may also be a right angle.

Solution: (1) When two right-angled sides are 3 and 4 respectively, the length of the third side is-=-= 5;

(2) When the hypotenuse is 4 and the right angle is 3, the length of the third side is-=-.

Example 2. The two sides of a right triangle are 6 and 8 respectively, so the radius of the circumscribed circle of this triangle is equal to _ _ _ _ _.

Analysis: this side of 8 can be regarded as both a right angle and a hypotenuse, so there are two possibilities for the hypotenuse of this right triangle to be 8 or 10, and then there are two possibilities for the radius of the circumscribed circle to be 4 or 5. A: Four or five.

Example 3. In rectangular ABCD, AB=5, BC= 12. If two circles centered on A and C are tangent, point D is inside circle C and point B is outside circle C, then the range of radius R of circle A is

Analysis: If the "tangency" condition is not well considered, there will be missing solutions.

Solution: Let the radius of circle C be r', then get 5 from the meaning of the question.

∴5<； 13-r & lt； 12, 1.

When circle A is inscribed with circle C, r-r'= 13.

∴5

So the range of R is 1.

Example 4. How many straight lines can be drawn when three points intersect on a plane?

Analysis: Because of the fixed thinking, when drawing three points, they are usually drawn on different straight lines, ignoring the situation that the three points are on the same straight line.

Correct answer: You can draw one or three straight lines when three points intersect on a plane.

Example 5. In the same plane, if the longest distance from point P to ⊙O is 8cm and the shortest distance is 2cm, then the radius of ⊙O is _ _ _ _ _.

Solution: Because there are two possibilities for the positional relationship between point P and ⊙O as shown in the figure.

∵AB is the diameter of⊙ O, Pb = 2cm, Pa = 8cm ∴ OA = OB =-(PA-Pb) = 3cm or OA = OB =-(Pb+PA) = 5cm, then the radius of⊙ O should be 5cm or 3cm.

Example 6. ⊙O has a diameter of 6cm. If the distance from point C to point O on straight line A is 3cm, then the positional relationship between straight line A and ⊙ O is _ _ _ _.

Solution: the problem only involves the distance from point C to the center of the circle, not the distance from the center of the circle to the straight line. As shown in Figure 2, there are two possibilities, so the positional relationship between straight lines A and ⊙O is tangent or intersecting.

Example 7. The radius ⊙ O is 5cm, and the chord AB∨CD, AB=6cm, CD=8cm. Find the distance between AB and CD.

Analysis: the relationship between AB and CD and the center of the circle in this question is not clear. Because the positions of AB and CD are uncertain, consider the case that the center of the circle is solved between two parallel chords and the center of the circle is outside the two parallel chords.

Solution: If the intersection o is a straight line OE⊥AB, the vertical foot is E, and CD intersects at point F, then ⊥ CD, AE = AB = 3, and OA is connected with OC. In Rt△AOE, OE =-=-4. In the same way, we can get of = 3.

∴ ef = OE+of = 7 or 4-3= 1

So the distance between AB and CD is 1cm or 7cm.

Example 8. ⊙O 1 and ⊙O2 intersect at point A and point B, with radius ⊙O 1 being 10, radius ⊙O2 being 17, and chord AB= 16.

Solution: For the problem of the intersection of two circles, students often only consider the situation that the centers of the two circles are on both sides of the common chord, that is, the situation of the diagram (1), and it is easy to miss the situation of the diagram (2), so the correct answer is O 1O2=2 1 or O 1O2=9.

Example 9. If the radius ⊙ O is 1cm, the chord AB=-cm and AC=-cm, then ∠BAC=___

Solution: Since the chords AB and CD may be on the same side or on the opposite side of the center, there are two possibilities, as shown in the figure. According to the vertical diameter theorem and the knowledge of solving right triangle, ∠ Cao = 45 and ∠ Bao = 30, so ∠ BAC = 15 or ∠ BAC = 75.