It is known that the equation of circle m is x ^ 2 + (Y-2) ^ 2 = 1, the equation of line L is x-2y = 0, the point P is on line L, the tangent PA and Pb of circle m are made through P, and the tangent points are a and B If the angle APB is 60 °, try to find the coordinates of point P

It is known that the equation of circle m is x ^ 2 + (Y-2) ^ 2 = 1, the equation of line L is x-2y = 0, the point P is on line L, the tangent PA and Pb of circle m are made through P, and the tangent points are a and B If the angle APB is 60 °, try to find the coordinates of point P

Let P (m, M / 2) link MP
Because ∠ APB = 60 °, so ∠ MPa = 30 °
In RT △ amp, MP = 2mA = 2
That is M & sup2; + (M / 2 - 2) & sup2; = 4
The solution is m = 0 or M = 8 / 5
So the coordinates of point P are (0,0) or (8 / 5,4 / 5)
It is known that the circle M: x2 + (y-4) 2 = 4, the equation of the straight line L is x-2y = 0, the point P is a moving point on the straight line L, the point P is the tangent PA and Pb of the circle, and the tangent points are a and B
Verification: the line AB must pass through a fixed point, and the coordinates of the fixed point can be obtained
Let P (a, b), then a-2b = 0,
If two tangent lines are drawn through the circle in P direction and the tangent points are a and B respectively, then the equation of line AB is
Ax + (B-4) (y-4) = 4
It is reduced to AX + (B-4) y-4b + 12 = 0
A * x + b * (y-4) - 4 (y-4) - 4 = 0,
Let y-4 = - 2x, and - 4 (y-4) - 4 = 0, the solution is x = 1 / 2, y = 3,
Therefore, the line AB is always over the fixed point (1 / 2,3)
It is known that the circle M: x2 + (y-4) 2 = 4, the equation of the line L is x-2y = 0, the point P is a moving point on the line L, the tangent PA and Pb of the circle are made through the point P, and the tangent points are a and B
prove:
Let P (a, b), then a-2b = 0,
Two tangent lines are drawn through the circle in P direction,
The tangent points are a and B respectively,
Then the equation of line AB is
ax+(b-4)(y-4)=4 ，
Simplify... Unfold
It is known that the circle M: x2 + (y-4) 2 = 4, the equation of the line L is x-2y = 0, the point P is a moving point on the line L, the tangent PA and Pb of the circle are made through the point P, and the tangent points are a and B
prove:
Let P (a, b), then a-2b = 0,
Two tangent lines are drawn through the circle in P direction,
The tangent points are a and B respectively,
Then the equation of line AB is
ax+(b-4)(y-4)=4 ，
It is reduced to AX + (B-4) y-4b + 12 = 0.
The results are as follows
a*x+b*(y-4)-4(y-4)-4=0 ，
Let y-4 = - 2x and - 4 (y-4) - 4 = 0,
The solution is x = 1 / 2, y = 3,
Therefore, the line AB is always over the fixed point (1 / 2,3).
The proposition is proved!
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It is known that the circle M: x ^ 2 + (y-4) ^ 2 = 4, the equation of the line L is x-2y = 0, the point P is the moving point on the line L, the point P is the tangent PA, Pb of the circle, and the tangent points are a, B
Verification: the circle passing through points a, P and M must pass through the fixed point, and the coordinates of all the fixed points can be obtained
It is proved that: obviously, the circle passing through a, P and M must pass through the fixed point m (0,2), because Ma ⊥ AP, so the center of the circle passing through a, P and M is the midpoint of MP, the diameter of the circle is MP, the MQ ⊥ straight line l passing through M, and the perpendicular foot is Q, then the circle passing through a, P and M must pass through the fixed point Q, set Q (2y0, Y0) (q is on the straight line L: x-2y = 0), and the slope of the straight line L: x-2y = 0
(3) Let P (2m, m), the midpoint Q of MP
Because PA is the tangent of circle m, the circle passing through points a, P and M is a circle with Q as the center and MQ as the radius,
So the equation is: (x-m) 2 +
It is reduced to: x2 + y2-2y-m (2x + Y-2) = 0, which is an identity of M, so x2 + y2-2y = 0 and (2x + Y-2) = 0,
Solution or
Therefore, the circle passing through a, P and M must be expanded through the fixed point (0, 2) or
(3) Let P (2m, m), the midpoint Q of MP
Because PA is the tangent of circle m, the circle passing through points a, P and M is a circle with Q as the center and MQ as the radius,
So the equation is: (x-m) 2 +
It is reduced to: x2 + y2-2y-m (2x + Y-2) = 0, which is an identity of M, so x2 + y2-2y = 0 and (2x + Y-2) = 0,
Solution or
Therefore, the circle passing through a, P and M must pass the fixed point (0, 2) or be folded
Find the equation of the straight line 8 / 5 of the line segment cut by the circle x ^ 2 + y ^ 2-2x-2y + 1 = 0 through the point m (4,4)
We need to explain it in detail
(x-1)²+(y-1)²=1
Center (1,1), radius r = 1
Chord length 8 / 5, radius = 1
So chord center distance = √ [1 & sup2; - (8 / 5 △ 2) & sup2;] = 3 / 5
Let the slope of the line be K
y-4=k(x-4)
kx-y+4-4k=0
Chord center distance is the distance from the center of a circle to a straight line
|k-1+4-4k|/√(k²+1)=3/5
|k-1|=√(k²+1)/5
square
25(k-1)²=k²+1
12k²-25k+12=0
k=4/3,k=3/4
So 4x-3y-4 = 0 and 3x-4y + 4 = 0
Given two circles x + y = 1, x + y-2x-2y + 1 = 0, find the equation of (1) the line where their common chord lies. (2) the chord length of the line where their common chord lies is cut by circle: (x-1) + (Y-1) = 25 / 4
(1) By connecting two equations and substituting x2 + y2 = 1 into another equation, we can get the straight line equation where the common chord is located, that is, x + Y-1 = 0 (2). From the meaning of the title, we can know that the distance from the center of the circle to the straight line is 2 / 2 root sign 2, and the radius is 5 / 2, so we can get 2 / 2 root sign 23 on the other side by Pythagorean theorem, so the chord length is root sign 23
Find the sum circle C: x2 + y2-2x = 0 C2: x2 + Y2 + 4Y = 0 find the tangent length of circle C1, C2
(x-1) ^ 2 + y ^ 2 = 1, x ^ 2 + (y + 2) ^ 2 = 4, center distance = root 5
The relationship between circle C1: x2 + y2 = 4 and circle C2: (X-5) + y2 = 16 is
with
Given that two circles C1: x2 + Y2 + 6x-4 = 0 and circle C2: x2 + Y2 + 6y-28 = 0, judge the position relationship of two circles. If the intersection requests the length (x + 3) ^ 2 + y ^ 2 = 13 of the common chord of two circles,
Are you a sophomore?
Man, you're all wrong.
The position relationship between circle C1: x ^ 2 y ^ 2 2x-3 = 0 and circle C2: x ^ 2 y ^ 2-4y 3 = 0 is?
X & # 178; + Y & # 178; + 2x-3 = 0 can be changed into (x + 1) &# 178; + Y & # 178; = 4. This is a circle whose center is at a (- 1,0) point and radius is 2. X & # 178; + Y & # 178; - 4Y + 3 = 0 can be changed into X & # 178; + (Y-2) &# 178; = 1. This is a circle whose center is at B (0,2) point and radius is 1
It is known that AB is the diameter of circle O, AC is the chord of circle O, and point D is the midpoint of arc ABC
(1) It is proved that: connecting a and E, because AB is diameter, de ⊥ AB is arc BD = arc be, ∠ DAB = ∠ eaboa = OE, so ∠ AEO = ∠ EAB = ∠ dabd is the midpoint of AC arc, arc ad = arc CD ∠ AED is the circumference angle of arc ad, and ∠ DAC is the circumference angle of arc CD, so ∠ AED = ∠ DAC ∠ BAC = ∠ DAC - ∠ DAB ∠ OED = ∠ AED - ∠ AEC - ∠ AEC
As shown in the figure, ABC is a broken chord of circle O, BC > AB, D is the midpoint of ABC arc, de ⊥ BC, and the perpendicular foot is e. if DC and DB are connected, DC ^ 2-dB ^ 2 = AB * BC
Let Da, AC, DC be DF = dB and f be on BC, then there is an isosceles triangle DFB
In the graph, there are RT triangle Dec and RT triangle DEB, which are obtained by Pythagorean theorem
CD^2=CE^2+DE^2,DB^2=DE^2+BE^2,
DC^2-DB^2=CE^2+DE^2-DE^2-BE^2=CE^2-BE^2
=（CE-BE)(CE+BE)=(CE-BE)*BC=BC*AB
So as long as ab = ce-be
In the circle, there are circle angles ∠ DFB = ∠ DBC = ∠ DAC, ∠ DCB = ∠ DAB, ∠ BCA = ∠ BDA,
∵ D is the midpoint of ABC arc, ∵ CD = ad, ∵ DCA = ∵ DAC,
∠DCA=∠DCB+∠BCA=∠DAC=∠DBC=DFB=∠DCB+∠CDF,∴∠BCA=∠CDF,
In this paper, we present a new method to solve this problem, that is, BDA = CDF, CD = ad, DF = dB, triangle CDF ≌ triangle ADB,
∴CF=CE-EF=CE-BE=AB.