As shown in the figure, PA and Pb are two tangent lines of circle O, a and B are tangent points, AC is the diameter of circle O, ∠ BAC = 35 ° and the degree of ∠ P is calculated

∵ PA, Pb are two tangent lines of circle O, a and B are tangent points,
∴∠PAO=90°,∠PBO=90°
∵ AC is the diameter of the circle O, ∵ BAC = 35 
∴∠BOC=2∠BAC=70°
∵∠P=360°-∠PAO-∠PBO-∠AOB=∠BOC
∴∠P=70°

As shown in the figure, PA and Pb are tangent lines of ⊙ o, points a and B are tangent points, AC is the diameter of ⊙ o, ﹤ BAC = 20 °, then the size of ⊙ P is______ °．

Connect BC, OB;
∵ PA and Pb are tangent lines of ⊙ o, and points a and B are tangent points
∴∠OAP=∠OBP=90°，
∵ AC is the diameter of ⊙ o,
∴∠ABC=90°；
∵∠BAC=20°，
∴∠C=70°，
∴∠AOB=2∠C=140°，
∴∠P=180°-∠AOB=40°；
So the answer is 40 degrees

As shown in the figure, ⊙ o diameter AB length is 10, chord AC length is 6, bisector of ⊙ ACB intersects ⊙ o in D, then CD length is () A. 7 B. 7 Two C. 8 Two D. 9

As shown in the figure, AB is the diameter of circle O, and the chord CD ⊥

I don't know what to say. Please be more complete

As shown in the figure, in the circle O, the chord AB = CD, the extension Ba, DC intersect at point P, e is a point on arc dB, and CE intersects BD at point F. it is proved that PA = PC

[the question is not complete, points E and F can not be used. First prove PA = PC]
prove:
∵AB=CD
ν arc AB = arc CD (equal chord and equal arc)
ν arc AB + arc AC = arc CD + arc AC
That is arc BC = arc ad
Ψ d = ∠ B (equal arc to equal angle)
∴PB=PD
∴PB-AB=PD-CD
PA = PC

As shown in the figure, AB is the diameter of ⊙ o, the extension lines of chords BD and Ca intersect at point E, and the extension line of EF perpendicular to Ba is at point F （Ⅰ）∠DEA=∠DFA；（Ⅱ）AB2=BE•BD-AE•AC．

It is proved that: (I) connecting ad, ∵ AB is the diameter of the circle,  ADB = 90 ° and

AB is the diameter of circle O, the extension lines of chords DA and Ba intersect at point P, and BC = PC. It is proved that ab = AP, arc BC = arc CD

According to the meaning of the question, there is an error in your question: the extension lines of the chords DA and Ba intersect at the point P. the extension lines of the chords DA and BC intersect at the point P. the proofs are as follows: connecting AC. ∵ AB is the diameter, ∵ AC ⊥ CB. ∵ BC = PC, ≌ RT ≌ ACP (RT ⊿ i.e. right triangle)

In the unit circle with radius 1, if the length of a chord AB is root 2, then the angle of chord AB to the center of the circle is?

90 degrees

In the unit circle with the origin as the center and radius as 1, the length of a chord AB is 3. The number of radians of the center angle α of AB is______ .

Because in the unit circle with radius 1 and the origin as the center, the length of a chord AB is 0
3，
So the half chord length is:
Three
So the size of half of the center angle is α
2，sinα
2=
Three
Two
1＝
Three
2，
∴α
2＝π
3，
The center angle of the circle is α = 2 π
3．
So the answer is: 2 π
3．

Given that the center of the circle with radius 2 is at the coordinate origin, two mutually perpendicular chords AC and BD intersect at point m (1, root 2), find the maximum and minimum area of ABCD!

Because, the two diagonals of the quadrilateral ABCD are vertical. The area is s, so s (ABCD) = 1 / 2 (AC * BD) now m (x, y) = m (1, √ 2) AC = 2 √ (r? - y) = 2 √ (4-2) = 2 √ 2, BD = 2 √ (r? - x?) = 2 √ (4-1) = 2 √ 3, so s (ABCD) = 1 / 2 (AC * b)

As shown in the figure, PA and Pb are two tangent lines of circle O, a and B are tangent points, AC is the diameter of circle O, ∠ BAC = 35 ° and the degree of ∠ P is calculated