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As shown in the figure, if there is a chord AB with a length of 2 times the root sign 3 in a circle with a radius of 2cm, then the center angle of the circle to which the chord is directed is ∠ AOB A 60 degrees B90 degrees C120 degrees D150 degrees
See the picture (see resources)
AB = 2 √ 3 D is the midpoint of AB, and ∠ ODB = 90 ° can be seen from the vertical diameter theorem
In RT △ ODB, OB = 2 dB = √ 3
sin∠DOB=√3/2
∠DOB=60°
The center angle of the circle to which the chord is aligned ∠ AOB = 120 °
Therefore, C is selected
If we don't know, we'll discuss it later
In a circle with radius 1, the degree of the angle of the center of the circle opposite to the chord with the root of 2 is
90°
If the radius of a circle is 2 cm, and the length of a chord in the circle is 2 times as long as the root sign is 3 cm, then the distance between the midpoint of the chord and the key point of the arc opposite the chord is equal to
The length of the chord is equal to √ 3. Connect the end point of the string with the center of the circle to form a right triangle. The distance between the chord center and the center is 1, and 2-1 = 1 is the distance between the midpoint of the chord and the midpoint of the arc
The radii are known to be 4 and 2 If the common chord length is 4, then the center distance between the two circles is______ .
In RT △ o1ac, o1c=
O1A2−AC2=
(2
2)2−22=2，
Similarly, in RT △ o2ac, O2C = 2
3，
∴O1O2=O1C+O2C=2+2
3．
There is another case, O1O2 = o2c-o1c = 2
3-2．
How much is 1 / 1 + root 2 + 1 / root 2 + root 3 + 1 / root 3 + root 4 + 1 / root 4 + root 5 + 1 / root 5 + root 6
11 / 2 plus 3 / 2 root sign 2 + 3 / 4 root sign 3 + 6 / 5 root sign 5 + root sign 6
Given that the radius of circle O is 1cm, chord AB = radical 3, AC = radical 2, find the degree of angle BAC Two possibilities. Explanation of drawing
Connect OA, ob, OC
∵ AB = radical 3,
∴∠OAB=30°
∵ AB = radical 2
∴∠OAC=45°
When o is inside ∠ BAC, ∠ BAC = 45 + 30 = 75 °
When o is outside of ∠ BAC, ∠ BAC = 45-30 = 15 °
Circle O1 intersects with circle O2 at two points a and B. the radius of circle O1 is 8 cm and that of circle O2 is 6 cm. If O1O2 = 10 cm, what is ab equal to
AB = 9.6cm
Circle O 1 and circle O 2 intersect at two points a and B. the radius of circle O 1 is 10 cm, the radius of circle O 2 is 8 cm, and the common chord length AB = 12 cm
In RT △ o1ad, O1A = 10cm, ad = 6cm, according to the Pythagorean theorem o1d = root (O1A · - AD) = 8cm.. similarly, BD = root (8 · - 6 ·) = 7cm. Therefore, O1O2 = o1d + o2d = (8 + 2 7) cm... Therefore, O1O2 = o1d + o2d = (8 + 2 roots 7) cm... That is, the length of O1O2 of the connecting heart line O1O2 is (8 8 + 2 7) cm... That is, the length of O1O2 of the connecting center line is (8 8 8 + 2 7) cm... That is, the length of O1O2 of the connecting center line is (8 8 8 + 2 7) cm (8 + 2) O2 is (8 + 2) length of O1O2 is (8 + 2+ 2 roots 7)
As shown in the figure, circle O1 intersects with circle O2 at points a and B, respectively connecting AB and O1O2. Verify ab ⊥ O1O2
Connect O1A, o1b, O2A, o2b,
∵ O1A = o1b (equal radius),
﹤ O1 is on the middle perpendicular line of AB (the points with equal distance to both ends of the line segment are on the middle perpendicular line of the line segment)
Similarly, ∵ O2A = o2b,
/ / O2 is on the vertical line of ab,
﹣ O1O2 is the perpendicular line of segment AB (two points determine a straight line)
∴AB⊥O1O2
As shown in the figure, the radius of circle O1 and circle O2 are equal to 1, O1O2 = 4. The moving point P is the tangent line PM and PN of circle O1 and circle O2 respectively (M and N are the tangent points) As shown in the figure, the radii of circle O1 and circle O2 are equal At 1, O1O2 = 4. The transition point P is the circle O1, O1O2 = 4 Tangent PM and PN of circle O2 (M and N are tangent points), Make | PM | = radical 2 | PN |. Try to establish a plane right angle seat And find the trajectory equation of the moving point P. take O1 as the origin of the coordinate system
P(x,y)
O1(0,0),O2(4,0)
PM^2/PN^2=(O1P^2-O1M^2)/(O2P^2-O2N^2)=2
[(x^2+y^2)-1]/[(x-4)^2+y^2-1]=2
(x-8)^2+y^2=33
O1(0,0),O2(-4,0)
(x+8)^2+y^2=33
RELATED INFORMATIONS
- 1. As shown in the figure, the radii of circles O1 and O2 are 5cm and 4cm respectively. The two circles intersect points a and B, and ab = 6cm. Find the length of the distance between the center of the circle and O1O2
- 2. It is known that circle O1. Circle O2 is inscribed at point P. the chord ab of circle O1 intersects circle O2 at two points of C. D, connecting PA.PC.PD Let Pb intersect with circle O2 at point E To prove (1) Pa * PE = PC * PD (2) if "circle O1. Circle O2 inscribed at point P" is changed to "circle O1. Circle O2 circumscribed to point P", other conditions remain unchanged, then whether the conclusion in (1) is tenable? Please explain the reason... For these mathematical problems, I have used up my integral! If I do not complete it on time, I will not be able to read and register
- 3. In circle O, diameter AB = 4, chord AC = 2, root sign 3. Chord ad = root 2, find the degree of arc CD
- 4. Take any point o on the line AB, and make OC and OD ray through the point O to make the angle cod = 90 degrees. When the angle AOC = 30 degrees, calculate the degree of angle BOD
- 5. Given that the radius of ⊙ o is OA = 10cm, chord AB = 16cm, P is a moving point on chord AB, then the shortest distance of OP is______ cm．
- 6. In circle O, the chords AB and CD intersect perpendicularly at the point E, AE = 5, CE = 1, be = 3. Find the radius of circle o
- 7. As shown in the figure, point C is a point on the circle, ⊙ o diameter AB is 10cm, the bisector of ⊙ ACB intersects ⊙ o in D. (1) if the chord AC is 6cm, find the length of BC and AD
- 8. As shown in the figure, AB is the diameter of ⊙ o, CD is the chord, CE ⊥ CD intersects AB with E, DF ⊥ CD intersects AB with F, proving that AE = BF
- 9. If CD ⊥ AB, prove en = n
- 10. In the circle O, the chord AB is parallel to CD, and it is proved that ad = BC (AD and BC intersect)
- 11. As shown in the figure, take the chord ab of ⊙ o as the edge, make a square ABCD outside the circle, and make the tangent DM and CN of ⊙ o through points D and C respectively, and make tangent DM of ⊙ o through points D and C respectively Cn, the tangent points are m and n (1) Verification, DM = cn (2) If AB = 2, DM = 2, root 2, find ⊙ o radius
- 12. As shown in the figure, in the two concentric circles with point o as the center, the chord ab of the large circle is the tangent line of the small circle, and the point P is the tangent point
- 13. As shown in the figure, AB is the diameter of circle O, AB is perpendicular to chord CD at point P, and P is the midpoint of radius ob, CD = 6, then the diameter of circle O is
- 14. As shown in the figure, in ⊙ o, C is If BC = 10 and Ce: EB = 3:2, find the length of ab
- 15. As shown in the figure, in the quadrilateral ABCD, ∠ B = ∠ ACD, ab = 6, BC = 4, AC = 5, CD = 7 and 1 / 2, find the length of AD rtrt
- 16. As shown in the figure, in diamond ABCD, e is the midpoint of AD, and the extension line of EF ⊥ AC crossing CB is at F Verification: AB and EF are equally divided
- 17. As shown in the figure, BC is known to be the diameter of ⊙ o, points a and F are on ⊙ o, ad ⊥ BC, perpendicular foot is D, BF crosses ad to e, and AE = be (1) Results: ab = AF; (2) If sin ∠ FBC = 3 5，AB═4 5. Find the length of AD
- 18. AB is the diameter of circle O, BC is the chord, OD ⊥ BC is at point E, intersection arc BC is at point D. please write five different types of correct conclusions If BC = 8, ed = 2, find the radius of circle o
- 19. It is proved that: 1, AE = BD 2, if AC is vertical be, prove AD + BD = CD × Radix 2
- 20. As shown in the figure, draw ⊙ o with ab side of positive △ ABC as the diameter, intersect AC and BC at points D and e respectively, known AB = 6cm, calculate the length of Arc de and the area of shadow part