P is a point outside the circle O, Pb is the tangent line of circle O, a and B are the tangent points, BC is the diameter of circle O, so AC is parallel to Po As the title

P is a point outside the circle O, Pb is the tangent line of circle O, a and B are the tangent points, BC is the diameter of circle O, so AC is parallel to Po As the title

Connect AB to Po to D, connect OA
Because BC is the diameter, the angle cab = 90
Namely: CA vertical ab
And triangle OAP is equal to triangle OBP
Angle AOP = angle BOP
OA＝OB
So OP is vertical ab
So AC / / op

As shown in the figure, the vertical bisector of chord AB intersects at point C and intersects chord AB at point D. It is known that ab = 24cm, CD = 8cm (1) Find the circle where the fragment is located (do not write the method, keep the drawing trace); (2) Find the radius of the circle in (1)

(1) The vertical bisector of chord AC and the vertical bisector of chord AB intersect at point O. take o as the center of the circle and the length of OA as the radius. O is the circle where the fragment is located, as shown in Fig
(2) When OA was connected, OA = x, ad = 12cm, OD = (X-8) cm,
According to the Pythagorean theorem, the following equations are formulated
x2=122+（x-8）2，
The solution is: x = 13
A: the radius of the circle is 13 cm

As shown in the figure, on the broken circular fragment, the vertical bisector of chord AB intersects the arc AB at point C and the intersecting chord AB at point D. AB = 8cm, CD = 2cm (1) Find the circle where the fragment is located (do not write the method, keep the drawing trace); (2) Find the radius of the circle in (1)

(1) The drawing is as follows,
         
(2) Let the radius of the circle p be r,
∵AB⊥CD，AB=8cm，CD=2cm，
∴AD=1
2AB=4cm，PD=r-2cm，
In Rt △ APD, AP2=AD2+DP2,
∴r2=42+（r-2）2，
The solution is r = 5,
The radius of ⊙ P is 5cm

If chord AB in circle O is the vertical bisector of radius, then the degree of arc ACB?

The degree of arc ACB is 240 degrees

As shown in the figure, the radius of circle O is a fixed length R, a is a certain point outside circle O, and P is any point on the circle. The vertical bisector l of line AP and the straight line OP intersect at point Q. when point P moves on the circle, the trajectory of point Q is () A. Ellipse B. Circle C. Hyperbola D. Straight line

∵ A is a certain point outside ⊙ o, and P is the last moving point of ⊙ o
The intersection line op of the vertical bisector of line AP is at point Q,
Then QA = QP, then qa-q0 = qp-qo = OP = R
That is, the distance difference between the moving point Q and the two fixed points o and a is a fixed value,
According to the definition of hyperbola, the locus of point q is a hyperbola with O and a as the focus and OA as the real axis
Therefore, C

Given the circle x ^ 2 + y ^ 2 = 16, fixed point a (2,0) if P is a moving point on the circle, the vertical bisector of AP intersects OP at R. find the trajectory equation of R

Because R is on the AP vertical bisector
So there was
AR=PR
Because or + PR = OP = 4
obtain
OR+AR=4
So the trajectory of R is an ellipse
2a=4
A=2
Focus (0,0) and (2,0)
2c=2
C=1
b^2=a^2-c^2=3
Center point (1,0)
So the trajectory equation is (x-1) ^ 2 / 4 + y ^ 2 / 3 = 1
Please click to select the satisfactory answer,

Circle B (x + 3) 2 + y2 = 16, a (3,0), P is any point on the circle, and Q is the intersection point of AP's perpendicular line and OP, so we can find the trajectory equation of Q Ladies and uncles

If Q (x0, Y0), P (x, y), then: OP: y = x * Y0 / x0 ① AP midpoint m [(x + 3) / 2, Y / 2] is brought in by KQM * KAP = - 1 and sorted out as follows: y ^ 2 + X ^ 2-2yy0-2xx0 + 6x0-9 = 0

On the circle (x + 2) ^ 2 + y ^ 2 = 25, a (- 2,0), B (2,0), move a point P on the circle, make a vertical bisector of BP, intersect AP with point m, and find the trajectory equation of M

If the midpoint of Pb is n, then the perpendicular crosses N and crosses AP at M
Then PM = MB
So am + BM = 5, so it is ellipse
Then it's good to ask

Given the fixed point a (radical 3,0) circle O: x ^ 2 + y ^ 2 = 4, P is the moving point on circle O, and the intersection radius of the middle perpendicular line of segment AP is op at m, the trajectory equation of point m is obtained

A(√3,0)
O:x^2+y^2=4,OP=r=2
M(x,y)
AM=PM
OP=OM+PM=OM+AM
2=√(x^2+y^2)+√[(x-√3)^2+y^2]
(x-0.5√3)^2+y^2/4=1

Given that the circle x ^ 2 + y ^ 2 = 4, and Q (radical 3,0), P is any point of the circle, then the locus m of the center perpendicular of PQ and OP is (o is the origin)

First of all, do these questions, I suggest you first draw pictures!
Basically draw a picture and you'll have half of the problem solved
Draw your own picture
Link MQ
Because the distance between the point on the vertical line of PQ and P and Q is equal;
So MP = MQ;
Because MP + om = r = 2;
So the M trajectory is an ellipse
And 2C = radical 3, a = 1 -------- MP + om = 2A = r = 2;
Then a ^ 2 = 1
b^2=a^2-c^2=1/4
It's good to replace the latter into the ellipse formula, but the x ^ 2 in the formula should be changed into (X-2 root sign 3) ^ 2
The final answer is:
(root of X-2 3) ^ 2 + 4Y ^ 2 = 1