It is known that the center of circle C is on the straight line y = 2x + 6 and passes through points a (0, - 1), B (- 2,3)

It is known that the center of circle C is on the straight line y = 2x + 6 and passes through points a (0, - 1), B (- 2,3)

(x-a)(x-a)+(y-b)(y-b)=c*c
2a+6=b
a*a+(b+1)(b+1)=c*c
(a+2)(a+2)+(b-3)(b-3)=c*c
result:a=-3 ,b=0,c*c=10
(x+3)(x+3)+y*y=10

In order to know that the circle C passes through a "3,2", B "1,6", and its center is on the straight line y = 2x, the equation of circle C is solved

Let the coordinates of the center of the circle be (a, b), and the distance from the two points is the same, then (A-3) ^ 2 + (b-2) ^ 2 = (A-1) ^ 2 + (B-6) ^ 2, then 2b-a = 6 is obtained by left-right expansion and cancellation. Because the center of the circle is on the straight line y = 2x, then B = 2A is substituted into the above formula to obtain a = 2, B = 4

It is known that the circle C passes through two points a (3,2) and B (1,6), and its center is on the straight line y = 2x It is known that the circle C passes through two points a (3,2) and B (1,6), and the center of the circle is on the straight line y = 2x. (1) if the straight line L passes through the point P (- 1,3) and is tangent to the circle C, find the equation of the straight line L

The center of the circle is on the vertical line of ab
AB slope -2, midpoint (2,4)
Vertical slope 1 / 2
x-2y+6=0
y=2x
Center C (2,4)
r=AC=√5
(x-2)²+(y-4)²=5
The distance from the center to the tangent is equal to the radius
y-3=k(x+1)
|2k-4+3+k|/√(k²+1)=√5
figure out
2x-y+5=0,x+2y-5=0

If the distance between the center of circle x2 + y2-2x-4y = 0 and the straight line X-Y + a = 0 is Two 2, then the value of a is______ .

The equation of a circle is reduced to the standard formula: (x-1) 2 + (Y-2) 2 = 5, so the coordinates of the center of the circle are (1,2)
Then the distance d from the center of the circle to the straight line X-Y + a = 0 is d = | 1 − 2 + a|
12+ (−1)2=
Two
2, namely | A-1 | = 1, simplify to get A-1 = 1 or A-1 = - 1, and get a = 2 or a = 0
So the value of a is 0 or 2
So the answer is: 0 or 2

As shown in the figure, in the plane rectangular coordinate system, the coordinates of points a and C are (- 1,0) (0, negative root sign 3) and point B is on the X axis As shown in the figure, in the plane rectangular coordinate system, the coordinates of points a and C are (- 1,0) (0, - 3) respectively, and point B is on the x-axis. The image of a certain quadratic function passes through points a, B and C, and its symmetry axis is a straight line x = 1. Point P is a moving point on the image of quadratic function below the line BC (point P does not coincide with point B and C). The passing point P is a parallel line of Y-axis and intersects BC with point F Find the analytic expression of the quadratic function? If the abscissa of point P is m, the length of line PF can be expressed by algebraic expression containing M? At this time, find the maximum area of △ P? sorry

1,2

As shown in the figure, in the plane rectangular coordinate system, three points a (B-2, root a), B (C, root a), C (a, a, root a), where a, B, C satisfy the relationship | A-2 | + (B-3) 2 + Radix 4-C = 0, point m moves from point a to the right along ray AB at the speed of 2 unit length per second, and point n moves from point B along suspected AB at the speed of 1 unit length per second. When line segment Mn = 1, calculate the area of △ OCM Trouble a seat, although there is no reward

a - 2 = 0, a = 2
b - 3 = 0, b = 3
4 - c = 0, c = 4
A (1, √ 2), B (4, √ 2), C (2, √ 2) are on the same straight line
T second: m (1 + 2T, √ 2), n (4 + T, √ 2)
Mn = 1 = 4 + T - (1 + 2t) = 1, t = 2, m (5, √ 2), △ OCM area = (1 / 2) (5-2) * √ 2 = 3 √ 2 / 2
or
Mn = 1 = 1 + 2T - (4 + T), t = 4, m (9, √ 2), △ OCM area = (1 / 2) (9-2) * √ 2 = 7 √ 2 / 2

In the plane rectangular coordinate system xoy, the line x + (M + 1) y = 2 - M and the straight line In the plane rectangular coordinate system xoy, the line x + (M + 1) y = 2 - M and the straight line MX + 2Y = - 8 are mutually perpendicular if and only if M=______________ . How do you calculate this? What's the procedure? Do you need a formula? It will be adopted in a short time,

x + (m+1)y = 2 - m
Finishing: y = (2 - M - x) / (M + 1) = - X / (M + 1) + (2 - M) / (M + 1), K1 = - 1 / (M + 1)
mx + 2y = -8
Finishing: y = (- 8 - MX) / 2 = - (MX) / 2 - 4, K2 = - M / 2
Mutually perpendicular K1 * K2 = - 1
[- 1/(m+1)]*（-m/2）=-1
m=-2/3

It is known that point a (m, 6) B (n, 1) is two moving points in the plane rectangular coordinate system xoy, where o is less than m and less than 3, connecting OA, OB.OA When the area triangle = 10, the parabola passes through two points a and B, and takes the Y axis as the symmetry axis, and finds the relationship of the quadratic function corresponding to the parabola

(1) Let ax denote the abscissa of point a, ay denote the ordinate of point a, and point B is similar, then AB? = (ay by) 2 + (AX BX) 2 = (6-1) 2 + (m-n) 2, and OA ⊥ ob, then ab ⊥ ob, ab ⊥ ob, ab ⊥ ob, ab ⊥ ob ⊥ then ab ⊥ ob ⊥ then ab ⊥ ob ⊥ then ab ⊥ ob ⊥ then ab

In the plane rectangular coordinate system xoy, points a (- 1, - 2), B (2,3), C (- 2, - 1) Find (1) the length of two diagonals of a parallelogram adjacent to line segments AB and AC (2) Let the real number t satisfy (vector ab-t * vector OC) * vector OC = 0, find t

1、
The diagonal lines ad, BC, intersect at point o
BC=4√2
E is the midpoint of B and C, e (0,1)
AD=2AE=2√10
2、
AB=(3,5) OC=(-2,-1)
(vector ab-t * vector OC) * vector OC = 0
(3+2t,5+t)(2,1)=0
11+5t=0
t=-11/5

In the plane rectangular coordinate system xOy, we know A (1,0), B (4,0), C (3,2); if we want to make △ ABD and △ ABC congruent, then the coordinate of point D is_

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