It is known that the equation of circle C is x2 + y2-8x-2y + 10 = 0, and the linear equation of the shortest chord passing through point m (3,0) is () A. x+y-3=0B. x-y-3=0C. 2x-y-6=0D. 2x+y-6=0

It is known that the equation of circle C is x2 + y2-8x-2y + 10 = 0, and the linear equation of the shortest chord passing through point m (3,0) is () A. x+y-3=0B. x-y-3=0C. 2x-y-6=0D. 2x+y-6=0

Circle x2 + y2-8x-2y + 10 = 0, that is, (x-4) 2 + (Y-1) 2 = 7, which means a circle with a radius equal to 7 and C (4,1) as the center. Obviously, point m (3,0) is inside the circle, so when the line is perpendicular to cm, the chord length is the shortest, so the slope of the line where the shortest chord is located is − 1kcm = − 11 − 04 − 3 = - 1
Delta ABC is an equilateral triangle, O is a point of delta ABC, OA = 5 ob = 4 OC = 3 try to find the degree of delta BOC
Rotate the triangle AOB 60 ° counterclockwise to the CDB
So triangle AOB is equal to BDC
∠OBD=60°
So ob = BD,
So triangle OBD is equilateral triangle
So od = ob = 4
CD=AO=5
In triangle cod
OC=3,OD=4.CD=5
So triangle OCD is right triangle
Therefore, COD = 90
Therefore, cob = 90 + 60 = 150
Take B as the original point, rotate the triangle BOC 60 degrees counterclockwise, O new position, P, C new position coincide with a
Then: AP = OC = 3, Pb = 4, ∠ BOC = ∠ APB
And BPO is equilateral triangle
∠BPO=60
AP^2+BP^2=3^2+4^2=5^2=AO^2
So: ∠ apo = 90
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It is known that the equation of circle C is x2 + y2-8x-2y + 12 = 0. The linear equation of the longest chord and the shortest chord of a point (3,0) in the circle is solved
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(x-4) ^ 2 + (Y-1) ^ 2 = 5, that is, the longest chord of the center (4,1) passing through the inner point of the circle is the diameter, that is, passing through the point (3,0) (4,1) the linear equation is y = x-3, and the shortest chord equation is y = - x + 3
The longest string is the straight line with diameter passing through the center of the circle and (3,0), and the shortest string is the one perpendicular to the line between the center of the circle and (3,0)
If a triangle can be formed with ABC = 135, AOC = 110, AOC = 135, AOC = 135, AOC = 135, AOC = 135, AOC = 135, AOC = 135, AOC = 135, AOC = 135, AOC = 135, AOC = 135, AOC = 135, AOC = 135, AOC = 135, AOC = 135, AOC = 135, AOC = 135
If the angle AOB remains unchanged, then when the angle BOC is equal to what degree, an equilateral triangle will be formed?
1. By rotating △ AOB clockwise 60 ° around B to make AB and BC coincide, it is obtained that △ BCO ≌ △ AOB ≌ △ BCO ≌ ob = O ′ B ∠ Bo ′ C = ∠ AOB = 110 ° OA = Co ′ ∫ OBO ′ = 60 °
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 coordinate of test point APB is 60 degrees
The equation of circle m is known to be 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, Pb of circle m is made through P, and the tangent points are a, B. if the angle APB is 60 °, then the distance from P to the center of circle is equal to 2. Let P (x0, Y0) | PM | = root sign [x0 ^ 2 + (y0-2) ^ 2] = 2x0-2y0 = 0, the solution is: x0 = 0, Y0 = 0x0 = 2 / 5, Y0 = 4 /
Let P (2m, m),
It can be seen from the question that MP = 1 / sin30 = 2, that is, (2m) 2 + (m-2) 2 = 4
The solution is: M = 0, M = 4 / 5, so the coordinates of the point P are p (0, 0) or P (8 / 5, 4 / 5).
As shown in the figure, ad is the angular bisector of △ ABC, AE is the height on the side of BC, ∠ B = 20 ° and ∠ C = 40 ° to find the degree of ∠ DAE
Ten
∠B=20°,∠C=40°
∠A=120°
∠BAE=60°
And because ∠ B = 20 ° and ∠ bea = 90 °
∠BAE=70°
∠DAE=∠BAE-∠BAE=70-60=10
10. Questioning: process
The chord length of the line passing through point P (2.3) cut by circle x2-2y + y2-3 = 0 is 2 √ 3
The chord length of a straight line passing through point P (2.3) cut by circle x2-2y + y2-3 = 0 is 2 √ 3. The equation for finding a straight line
The circle equation is x ^ 2 + (Y-1) ^ 2 = 4, the center of the circle is (0,1), and the radius is 2
If the chord length is 2 √ 3, the distance from the center of the circle to the chord = radius ^ 2 - (√ 3) ^ 2 = 4-3 = 1
If the straight line x = 2 passes through the point P (2,3), the chord length cut by the circle is 2 √ 3
Let Y-3 = K (X-2), that is, kx-y-2k + 3 = 0
The distance from the center of the circle (1,0) to the straight line kx-y-2k + 3 = 0 = [- K + 3] / √ (k ^ 2 + 1) = 1, the solution is k = 4 / 3
The straight line is (4 / 3) x-y-8 / 3 + 3 = 0, that is 4x-3y + 1 = 0
Therefore, the equations of the straight line are x = 2 and 4x-3y + 1 = 0
The equation of circle is X & # 178; + (Y-1) &# 178; = 4, and the radius is 2
Let the linear equation over p be Y-3 = K (X-2)
Then the distance from P to the center of the circle is 1
(2k-2)²=k²+1
K=
Let the line passing through point P (2.3) be Y-3 = K (X-2)
The standard equation of circle: x2 + (Y-1) 2 = 4, Center (0,1), radius 2
According to Pythagorean theorem, the chord center distance is 1
(-2k+2)2=1+k2
The solution is K1 = (4 + √ 7) / 3, K2 = (4 - √ 7) / 3
Through P line y = K (X-2) + 3
kx-y-2k+3=0
x^2-2y+y^2-3=0
X ^ 2 + (Y-1) ^ 2 = 4 center o '(0,1) radius 2
Chord length 2 √ 3
Distance from o 'to straight line d = √ 4 - (2 √ 3 / 2) ^ 2 = 1
d=|k*0-1-2k+3|/√(k^2+1)=1
(k-2)^2=k^2+1
4k=3
k=3/4
Straight line y = (3 / 4) (X-2) + 3
Let the linear equation be Y-3 = K (X-2)
It is reduced to the general formula KX - y + (3-2k) = 0
The equation of the circle is reduced to the standard formula X & # 178; + (Y-1) &# 178; = 4
Obviously, the radius of the center (0,1) r = 2
Make a vertical line through the center of the circle. According to the vertical diameter theorem, the vertical foot is the middle point of the string.
The chord center distance D is expressed by the distance formula from point to line
Note: radius, chord center distance and half chord form right triangle,
Start with Pythagorean
Let the linear equation be Y-3 = K (X-2)
It is reduced to the general formula KX - y + (3-2k) = 0
The equation  -  -  -  -  -  - 178;  -  - 178;  -  - 178;  -  - 178
Obviously, the radius of the center (0,1) r = 2
Make a vertical line through the center of the circle. According to the vertical diameter theorem, the vertical foot is the middle point of the string.
The chord center distance D is expressed by the distance formula from point to line
Note: radius, chord center distance and half chord form right triangle,
Using Pythagorean theorem formulation, we can solve K and bring back the linear equation.
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Let the equation of straight line be y = KX + B;
The center of circle x2-2y + y2-3 = 0 (0,1), radius 2;
If the chord length is 2 √ 3, the distance from the center of the circle to the straight line is √ (2 ^ 2 - √ 3 ^ 2) = 1;
1=|k*0-1+b|/√(k^2+1)
k^2+1=b^2-2b+1
3=2k+b
The solution is b = (1 + 2 √ 7) / 3, k = (4 - √ 7) / 3
Or B = (1-2 √ 7) / 3, k = (4 + √ 7) / 3
As shown in the figure, in △ ABC, ad is the height on the edge of BC, AE is the bisector of ∠ BAC, ∠ B = 42 ° and ∠ DAE = 18 ° to find the degree of ∠ C
∵ ad is the height on the edge of BC, ∵ B = 42 °, ∵ bad = 48 °, ∵ DAE = 18 °, ∵ BAE = ∵ bad - ∵ DAE = 30 °, ∵ AE is the bisector of ∵ BAC, ∵ BAC = 2 ∵ BAE = 60 °, ∵ C = 180 ° - ∵ B - ∵ BAC = 78 °
Given that the line y = x + B and the circle x ^ 2 + y ^ 2 + 2x-2y + 1 = 0, if the line and the circle are tangent, we can find the equation of the line
First, the equation of a circle is reduced to a standard form
（x+1)²+(y-1)²=1
So the center of the circle is (- 1,1) and the radius is 1
So if the line y = x + B is tangent to the circle, then the distance from the center of the circle to the line should be equal to 1
Turn the equation of a straight line into
x-y+b=0
Thus ∣ - 1-1 + B ∣ = √ 2
That is ∣ B-2 ∣ = √ 2
So B = 2 + √ 2, or B = 2 - √ 2
If we go back to the original equation, there will be
Y = x + 2 + √ 2, or y = x + 2 - √ 2
6X ^ 2-xy-y ^ 2 + 5x-5y-4, and factorization in the range of rational number
6x²-xy-y²+5x-5y-4
=(2x-y)(3x+y)+4(2x-y)-(3x+y)-4
=(2x-y)(3x+y+4)-(3x+y+4)
=(2x-y-1)(3x+y+4)
=(2x-y-1)*(3x+y+4)