Given that the circle C (x + radical 3) ^ 2 + y ^ 2 = 16, point a (radical 3,0) q is a moving point on the circle, the vertical bisector of AQ intersects CQ at point m, and the trajectory equation of M is e Find the equation of E. urgent! The area of the l-intersection locus e of the straight line passing through the point P (1,0) at two different points a, B and △ AOB is 4 / 5

∵ the vertical bisector of AQ intersects CQ at point M
∴|MA|=|MQ|
∴|MA|+|MC|=|MQ|+|MC|=|CQ|=R=4
According to the definition of ellipse: the sum of distances from two fixed points on the plane is the locus of a point with constant value (2a)
﹤ e is an ellipse
2a=4,c=√3
∴b²=a²-c²=1
∴E：x²/4 + y²=1

As shown in the figure, PA and Pb tangent o to points a, B, and the radius of circle O is 3, ∠ APB = 60 °, connect AB, cross OP to point C, and find the length of Po, PA, AB, OC

Connect OA
∵ PA, Pb cut ⊙ o at point a, B,
∴∠OAP=90°，∠APO=1
2∠APB=30°，
∴OP=2OA=2
3，PA=
3OA=3，∠AOP=60°
∵ PA, Pb cut ⊙ o at point a, B,
∴PA=PB，
And ∵ BPA = 60 °,
ν Δ ABP is an equilateral triangle,
∴AB=PA=3，
∵∠AOP=60°
∴OC=OA•cos60°=3
2．

As shown in the figure, AB is the diameter of circle O, ad, BC, CD are tangent lines, a, B, e are tangent points, and CO ⊥ do is proved

Proof: connect OE
[proof 1] the length of two tangent lines from a point outside the circle is equal, that is, ad = De, CE = CB
∵ ad = De, OA = OE = radius, OD = OD
∴⊿DAO≌⊿DEO（SSS）
∴∠AOD=∠EOD
∵CE=CB,OB=OE,OC=OC
∴⊿CBO≌⊿CEO（SSS）
∴∠BOC=∠EOC
∴∠DOE+∠COE=∠AOD+∠BOC=180º÷2=90º
That is, Doc = 90 ° and CO ⊥ do
[syndrome 2]
Tangent vertical radius outer end, i.e. ∠ oad = ∠ OED = ∠ OEC = ∠ OBC = 90 degrees
∵AO=EO,OD=OD,∠OAD=∠OED=90º
∴Rt⊿OAD≌RT⊿OED（HL）
Similarly: RT ⊿ OEC ≌ RT ⊿ OBC (HL)
(the following is the same as the method of proof 1, omitted)

As shown in the figure, AB is the diameter of ⊙ o, C is the point on ⊙ o, ad is perpendicular to the tangent of point C, and the perpendicular foot is d (1) Verification: AC bisection ∠ bad; (2) If AC = 2 5, CD = 2, find the diameter of ⊙ o

(1) It is proved that: as shown in the figure, connect OC, ∵ DC cut ⊙ o in C,  OC ⊥ CF,  ADC =  OCF = 90 °, ad ∥ OC,  DAC = ∠ OCA, ? OA = OC, ? OAC = ∠ OCA, that is, AC bisection ? bad. (2) connect BC. ? AB is the diameter,  ACB = 90 ° = ∠ ADC, ? DAC

As shown in the figure, it is known that AB is the diameter of ⊙ o, AC is the chord, and bisection ⊥ bad, ad ⊥ CD, and the perpendicular foot is d (1) It is proved that CD is ⊙ o tangent; (2) If the diameter of ⊙ o is 4 and ad = 3, find the degree of ⊙ BAC

(1) It is proved that: connecting OC, ∵ OA = OC,  OCA = ∠ OAC. ? AC bisection  bad,  BAC = ∠ CAD. ? OC ⊥ ad. and ? ad ⊥ CD, ? OC ⊥ CD. ? CD is tangent of ⊙ O. (4 points) (2) connect BC, ? AB is diameter,  BCA = 90 °

As shown in the figure, AB is the diameter of ⊙ o, AC is the chord, CD is the tangent of ⊙ o, C is the tangent point, and ad ⊥ CD is at point D （1）∠AOC=2∠ACD；（2）AC2=AB•AD．

It is proved that: (1) ∵ CD is tangent of ⊙ o,  OCD = 90 °,
That is ∠ ACD + ∠ ACO = 90 °. ① (2 points)
∵OC=OA，∴∠ACO=∠CAO，
Ψ AOC = 180 ° - 2 ∠ ACO, that is ∠ AOC + 2 ∠ ACO = 180 °,
Divide both sides by 2 to get: 1
(2) ACO (2 °) (AOC) = 4
From ①, ②, we get: ∠ acd-1
(5 points)
(2) As shown in the figure, connect BC
∵ AB is the diameter,  ACB = 90 °. (6 points)
In RT △ ACD and RT △ ABC,
∵∠AOC=2∠B，
∴∠B=∠ACD，
Rt △ ACD ∽ Rt △ ABC, (8 points)
∴AC
AB＝AD
AC, i.e. ac2 = ab · ad. (9 points)

As shown in the figure, it is known that AB is the chord of ⊙ o, C is the point above ⊙ o, ∠ C = ∠ bad, and BD ⊥ AB is at B (1) It is proved that ad is tangent of ⊙ o; (2) If the radius of ⊙ o is 3, ab = 4, find the length of AD

(1) It is proved that: as shown in the figure, connect AO and extend the intersection ⊙ o at point E, and connect be, then ? Abe = 90 degrees, ? EAB + ∠ e = 90 degrees. ?? e = C, ? C = ∠ bad, ? EAB + ∠ bad = 90 °. Ad is the tangent of ⊙ O. (2) from (1) we can see that  Abe = 90 degrees, diameter AE = 2ao = 6, ab = 4, ? be = AE2 − a

As shown in the figure, AB is the diameter of ⊙ o, C is a point on ⊙ o, the tangent lines of Da and passing point C are perpendicular to each other, and the perpendicular foot is d. if ∵ DAB = 70 °, calculate the degree of ⊙ DAC

Connect OC,
∵ CD is tangent,
∴OC⊥CD．
∵AD⊥CD，
∴AD∥CO，
∴∠1=∠3．
∵∠2=∠3，
∴∠1=∠2．
∵∠DAB=70°，
∴∠DAC=35°．

As shown in the figure, AB is the diameter of circle O, C is a point on circle O, the tangent lines of AD and passing point C are perpendicular to each other, and the perpendicular foot is d. (1) explain the bisection angle dab of AC; (2) if the knot is connected, the tangent line between AD and point C is perpendicular to each other "AC bisection angle DAB" is taken as the condition of the title, which shows that the tangent lines of AD and passing point C are perpendicular to each other. (3) if under the condition of (2), ad = 4, ab = 5, try to find the length of AC

One
Connect BC,
The tangent line is CD
Ψ DCA = ∠ B (chord tangent angle is equal to the circumferential angle of the circle to which the arc is clamped)
∵ AB is the diameter
∴∠ACB=90°,
∴∠CAB+∠B=90°,∠DCA+∠DAC=90°
Ψ DAC = ∠ cab (the remainder of an equal angle is equal)
That is, AC bisection ∠ DAB
Two
∵∠DAC=∠CAB,∠DCA=∠B,∠CAB+∠B=90°
﹤ DCA + ∠ DAC = 90 ° (equivalent substitution)
That is, the tangent lines of AD and passing point C are perpendicular to each other
Three
∵∠DCA=∠B,∠DAC=∠CAB
∴△DAC∽△CAB
∴AD/AC=AC/AB
That is, AC 2 = ad * AB = 20
∴AC=2√5

It is known that: as shown in the figure, AB is the diameter of circle 0, point C is on circle 0, tangent lines of AD and passing point C are perpendicular to each other, and the perpendicular foot is d. verification: AC bisection ∠ DAB

prove:
Connect OC
∵ CD is the tangent of circle o
∴OC⊥CD
∵AD⊥CD
∴AD//OC
∴∠DAC=∠ACO
∵OA=OC
∴∠ACO=∠CAO
∴∠DAC=∠CAO
That is, AC bisection ∠ DAB