When m is a value, the equation about X, y, (2m & # 178; + m-1) x & # 178; + (M & # 178; - M + 2) y & # 178; + m + 2 denotes a circle

When m is a value, the equation about X, y, (2m & # 178; + m-1) x & # 178; + (M & # 178; - M + 2) y & # 178; + m + 2 denotes a circle

First, the coefficients of x ^ 2 and Y ^ 2 must be equal
That is 2m ^ 2 + M-1 = m ^ 2-m + 2
The result is: m ^ 2 + 2m-3 = 0
(m+3)(m-1)=0
m=-3,1
When m = - 3, the equation is: 14x ^ 2 + 14y ^ 2-1 = 0, which is a circle;
When m = 1, the equation becomes: 2x ^ 2 + 2Y ^ 2 + 3 = 0, no solution, not circle
So we can only take M = - 3
The line L passing through the point m (2,2) is tangent to the circle (x-1) 2 + y 2 = 1. The equation of line L is obtained
When the slope of a straight line does not exist, the tangent equation is x = 2; when the slope of a straight line exists, let the tangent equation be Y-2 = K (X-2), that is, kx-y-2k + 2 = 0. The distance from the center of a circle (1,0) to the tangent is equal to the radius, and the solution is k = - 34. The tangent equation is 3x + 4y-14 = 0
Given the circle x2 + Y2 + KX + 2Y + K2 = 0, when the area of the circle is the maximum, the center coordinate is ()
A. （0，-1）B. （1，-1）C. （-1，0）D. （-1，1）
According to the meaning of the topic, the center coordinate of the circle is (- K2, - 1), and the radius of the circle is r = 124 − 3k2. In order to maximize the area of the circle, that is, the radius r of the circle is the maximum value, so when k = 0, r takes the maximum value of 1, and the center coordinate of the circle is (0, - 1)
Known ellipse: X & # 178 / 4 + Y & # 178; = 1. Straight line x + y = 1 intersects ellipse at A.B, and straight line y = KX (k > 0) intersects ellipse at C.D
Finding the maximum acbd area of quadrilateral
Ellipse: X & # 178 / / 4 + Y & # 178; = 1. The line x + y = 1 intersects the ellipse at two points a and B. the solution is a (0,1) B (8 / 5, - 3 / 5) AB = 8 / 5 × √ 2. When the line y = KX (k > 0) and the line x + y = 1 are perpendicular to each other, the area of quadrilateral ABCD is the largest. The analytical expression of the line y = KX (k > 0) is y = x ellipse: X & # 178 / / 4 + Y & #
Known ellipse: X & # 178 / 4 + Y & # 178; = 1. The line x + y = 1 intersects the ellipse at two points a and B, and the line y = KX
Ellipse: X & # 178 / / 4 + Y & # 178; = 1. The line x + y = 1 intersects the ellipse at two points a and B. the solution is a (0,1) B (8 / 5, - 3 / 5) AB = 8 / 5 × √ 2. When the line y = KX (k > 0) and the line x + y = 1 are perpendicular to each other, the area of quadrilateral ABCD is the largest. The analytical expression of the line y = KX (k > 0) is y = x ellipse: X & # 178 / / 4 + Y & #
If x-2y + 5 = 0 and circle x ^ 2 + y ^ 2 = 8 intersect at two points a and B, then the line AB = how much
Method 1 circle x ^ 2 + y ^ 2 = 8, the center of circle is O (0,0), radius r = 2 √ 2, the distance from the center of circle to the straight line AB d = 5 / √ 5 = √ 5 | ab | / 2 = √ (R & # 178; - D & # 178;) = √ 3 | ab | = 2 √ 3 method 2 straight line x-2y + 5 = 0 and circle x ^ 2 + y ^ 2 = 8 are X & # 178; + (x + 5) &# 178; + 4 = 8, that is 5x & # 178; + 10x-7 = 0, let a (x1, Y1), B
The line X - 2Y + 5 = 0 intersects the circle x 2 + y 2 = 8 at two points a and B
X-2Y+5=0
X2+Y2=8
(2Y-5)^2+Y^2=8
5y^2-20y+17=0
y1+y2=4 y1*y2=17/5
y1-y2=√[(y1+y2)^2-4y1*y2]
=2/5*√15
tona=k=1/2
sina=√5/5
The length of string AB = | y1-y2 | / Sina = 2 / 5 * √ 15 / √ 5 / 5 = 2 √ 3
2√3
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, so the equation of the line where the shortest chord passes through point m (3,0) is y-0 = - (x-3), that is, x + Y-3 = 0, so select a
Given the circle x ^ 2 + y ^ 2-8x-2y + 12 = 0, find the linear equation of the longest and shortest chord of a point P (3,0) in the circle
Circle: (x-4) ^ 2 + (Y-1) ^ 2 = 5
So the center O (4,1)
The line of the longest chord is the line of Po: y = x-3
The line where the shortest chord is located is the line perpendicular to Po: y = - x + 3
Know the circle x ^ 2
Circle: (x-4) ^ 2 + (Y-1) ^ 2 = 5
So Center (4,1), radius = root 5
Draw a figure, the shortest chord passing through point P, perpendicular to (circle center O and line segment of point P), draw a typical right triangle (vertical bisection, and radius).
The longest string is the equation of two points. The slope of OP is k = 1, and the line is y = x-3
The slope of the shortest chord = - 1, the line is y-0 = - 1 (x-3), that is y = - x + 3... Expansion
Circle: (x-4) ^ 2 + (Y-1) ^ 2 = 5
So Center (4,1), radius = root 5
Draw a figure, the shortest chord passing through point P, perpendicular to (circle center O and line segment of point P), draw a typical right triangle (vertical bisection, and radius).
The longest string is the equation of two points. The slope of OP is k = 1, and the line is y = x-3
The slope of the shortest chord = - 1, and the straight line is y-0 = - 1 (x-3), that is, y = - x + 3
It is known that the circle M: x2 + (Y-2) 2 = 1, the line L: x-2y = 0, the point P is on the line, the tangent PA and Pb of the circle m are made through the point P, and the tangent point is a
It is known that the circle M: x2 + (Y-2) 2 = 1, the straight line L: x-2y = 0, the point P is on the straight line, the tangent PA and Pb of the circle m are made through the point P, and the tangent points are a and B
(1) If ∠ APB = 60 °, try to find the coordinates of point P
(2) If the coordinate of point P is (2,1), a straight line passing through point P intersects with circle m at two points c and D, and when CD = root 2, the CD equation of the straight line is obtained.
(3) Verification: the circle passing through a, P and M must pass through the fixed point, and the coordinates of all the fixed points must be obtained
(1) Then | MP | = {[2 (Y0) - 0] ^ 2 + [(Y0) - 2] ^ 2} = 2,5 (Y0) ^ 2-4 (Y0) = 0, Y0 = 0 or Y0 = 4 / 5p
One
(1) Let P (2m, m)=
1sin30 ° = 2, that is, (2m) 2 + (m-2) 2 = 4 (3 points)
The solution is: M = 0, M = 0=
So the coordinate of the point P is p (0, 0) or P（
85，
45)．　 　　… (6 points)
(2) Let P (2m, m), the midpoint of MP Q (m,
M2 + 1), because PA is the tangent of circle M
So the circle passing through a, P and M is a circle with Q as the center and MQ as the radius
(1) Let P (2m, m)=
1sin30 ° = 2, that is, (2m) 2 + (m-2) 2 = 4 (3 points)
The solution is: M = 0, M = 0=
So the coordinate of the point P is p (0, 0) or P（
85，
45)．　 　　… (6 points)
(2) Let P (2m, m), the midpoint of MP Q (m,
M2 + 1), because PA is the tangent of circle M
So the circle passing through a, P and M is a circle with Q as the center and MQ as the radius,
So the equation is: (x-m) 2 + (y)-
m2-1)2=m2+(
m2-1)2… (9 points)
It is reduced to: x2 + y2-2y-m (2x + Y-2) = 0,
So x2 + y2-2y = 02x + Y-2 = 0 ﹥ 8203; the solution is x = 0y = 2 ﹥ 8203; or X=
45y=
25, i.e. (0, 2) and (45,
25）．… (14 points) put it away
Obviously, we can get a tangent position (0, 2)
And the slope of OP is 2
Then the slope of AB is - 1 / 2
So the equation of AB is y = - X / 2 + 2
。。。 。。。 Very speechless..
In that case,
OP=3√5
Draw a circle with P as the center and PA as the radius
Then PA ^ 2 = OP ^ 2-oa ^ 2 = 45-4 = 41
So the circle is
(x-3)^2+(y-6)^2=41
Subtract circle... Expand
Obviously, we can get a tangent position (0, 2)
And the slope of OP is 2
Then the slope of AB is - 1 / 2
So the equation of AB is y = - X / 2 + 2
。。。 。。。 Very speechless..
In that case,
OP=3√5
Draw a circle with P as the center and PA as the radius
Then PA ^ 2 = OP ^ 2-oa ^ 2 = 45-4 = 41
So the circle is
(x-3)^2+(y-6)^2=41
The equation of subtracting circle
We get - 6x + 9-12y + 36 = 41-4
3x + 6y-4 = 0
twelve million three hundred and forty-five thousand three hundred and twenty-one