As shown in the figure, lead the tangent PA, Pb of the circle through a point p outside ⊙ o with radius of 6cm, connect Po, intersect ⊙ o at F, make tangent of ⊙ o through F, intersect PA and Pb at D and e respectively, if Po = 10cm, ﹤ APB = 40 ° Find: (1) the circumference of △ ped; (2) the degree of ∠ doe

As shown in the figure, lead the tangent PA, Pb of the circle through a point p outside ⊙ o with radius of 6cm, connect Po, intersect ⊙ o at F, make tangent of ⊙ o through F, intersect PA and Pb at D and e respectively, if Po = 10cm, ﹤ APB = 40 ° Find: (1) the circumference of △ ped; (2) the degree of ∠ doe

As shown in the figure on the right
(1) Connect Ao, then OA ⊥ PA, PA=
PO2−OA2＝
102−62=8，
∵ PA, Pb are tangent lines, a, B are tangent points, EF, EB, DF, DA are tangent to ⊙ o,
∴PA=PB，DA=DF，FE=BE，
The circumference of △ ped = PE + EF + FD + PD = PA + Pb = 2PA = 16 (CM),
The circumference of △ PED is 16 cm;
(2) According to the property of tangent length, we know that: ∠ AOD = ∠ DOF, ∠ EOF = ∠ EOB,
∴∠DOE=1
2∠AOB=1
2（180°-∠APB）=1
2（180°-40°）=70°．

As shown in the figure, PA and Pb are two tangent lines of ⊙ o with radius of 1, points a and B are tangent points respectively, ﹤ APB = 60 °, OP and chord AB intersect at point C, and ⊙ o intersect point D. the area of shadow part is______ (the result retains π)

∵ PA and Pb are two tangent lines of ⊙ o with radius of 1,
⊥ PA, ob ⊥ Pb, Op ⊥ APB,
And ∠ APB = 60 °,
∴∠APO=30°，∠POA=90°-30°=60°，
And ∵ OP divides AB vertically,
∴△AOC≌△BOC，
∴S△AOC=S△BOC，
﹣ s shadow part = s fan-shaped oad = 60 π × 12
360=π
6．
So the answer is π
6．

PA and Pb are tangent lines of the center O of the circle, and the tangent points are a and B. given the angle P = 60 degrees and OA = 3, then the arc length of angle AOB is opposite to

∵ PA Pb is the tangent of circle O, ᙽ PA is perpendicular to Ao, Pb is perpendicular to ob
If the angle APB = 60 °, the angle AOB = 360-60-2 * 90 = 120 ° = 360 ° / 3,
The arc length corresponding to the angle AOB is 2 π R / 3 = 2 π

If PA is the tangent of circle O, a is the tangent point, Po intersects circle O at point B, PA = 4, OA = 3, then Pb =?

Connect ob, then ob ⊥ Pb,
In RT △ POB,
OB=OA=PO-AP=3,PO=5,
∴PB=PO2-OB2=52-32=4．

The radius of the circle O is known to be 1, PA and Pb are the two tangent lines of the circle. A and B are two tangent points. What is the minimum value of (vector) Pa × (vector) Pb? A, - 4 +

When PA ⊥ Pb, the vector product has the minimum value, which is the vector Po. - 3 + 2 radical sign 2

As shown in the figure, PA and Pb are tangent lines of ⊙ o, and the tangent points are a and B respectively, and ∠ APB = 50 ° and point C is a point on the arc ab (1) What do you think is the relationship between PA and Pb size? And explain the reasons; (2) Try to find the degree of ∠ ACB; (3) If point C is in the inferior arc AB, what is the degree of ∠ ACB?

(1) PA = Pb. The reasons are as follows: ∵ PA, Pb is the tangent line of ⊙ o, and the tangent points are a, B, and ᙽ PA = Pb; (2) connect OA and ob, as shown in the figure,

3. as shown in the figure, PA and PB respectively cut ⊙ O at A and B, and point C is a point on the superior arc, and the degree of ⊙ APB is () if ACB=60 °

60 ° process in a moment

Example 2. As shown in the figure, PA and Pb are respectively tangent to circle O at a and B, and intersect with tangent lines of circle O at C and D respectively. Known PA = 7cm, (1) calculate the circumference of △ PCD. (2) if ∠ P = 46 °, calculate the degree of ∠ cod

Test point: tangent length theorem. Analysis: because Da, DC, BC are all tangent lines of ⊙ o, according to the tangent length theorem, the perimeter of ⊙ PCD can be converted into the length of PA and Pb, and then the solution can be made. As shown in the figure, let the tangent point of DC and ⊙ o be e; ∵ PA, Pb are tangent lines of ⊙ o, and the tangent points are a and B;  PA = Pb = 7cm; similarly, we can get

As shown in the figure: PA, Pb cut ⊙ o at a, B, tangent intersection PA, Pb at D, e, PA = 8cm, then the circumference of ⊙ PDE is______ cm．

∵ PA, Pb cut ⊙ o to a, B, De to ⊙ o to C,
∴PA=PB=8，CD=AD，CE=BE；
The circumference of △ PDE = PD + PE + CD + CE = 2PA = 16 (CM)

Two tangent lines PA and Pb are introduced from the moving point P to the circle x2 + y2 = 1, and the tangent points are a and B respectively, ∠ APB = 60 °, then the trajectory equation of the moving point P is______ .

If the coordinates of point P are (x, y), then | Po|=
x2+y2
∵∠APB=60°
∴∠AP0=30°
∴|PO|=2|OB|=2
Qi
x2+y2=2
That is, X2 + y2 = 4
So the answer is: x2 + y2 = 4