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Given that the circle m is tangent to the x-axis, the center of the circle is on the straight line x-2y = 0, and the maximum distance from the moving point on the circle m to the y-axis is 3, find the square of the circle M
Let the center of the circle be (m, n) and the radius be r
m-2n=0,m=2n
r=|n|
|m|+r=3
Then | 2n | + | n | = 3
n=±1
When n = 1, M = 2, r = 1, the circle is
(x-2)^2+(y-1)^2=1
When n = - 1, M = - 2, r = 1, the circle is
(x+2)^2+(y+1)^2=1
When the distance between the center of circle C and the line KX + y + 4 = 0 is the largest, the value of K is?
The straight line KX + y + 4 = 0 can be changed into: y = - kx-4, constant crossing point m (0, - 4),
The circle x ^ 2 + y ^ 2 + 2x-2y + 1 = 0 can be changed into: (x + 1) ^ 2 + (Y-1) ^ 2 = 1, and its center C (- 1,1),
When the line KX + y + 4 = 0 is perpendicular to the line cm, the distance from the center of circle C to the line KX + y + 4 = 0 is the largest,
Because the slope of the line cm = (1 + 4) / (- 1-0) = - 5,
So: k = 1 / 5
It is known that the circle C passes through two points P (- 1, - 3). Q (2,6), and the center of the circle is on the straight line x + 2y-4 = 0. The equation of the straight line L is (k-1) x + 2Y + 5-3k = 0
Find the shortest chord length of the straight line L cut by the circle C? Time is tight, it's going to start tonight
Linear PQ equation: (6 + 3) / (2 + 1) = (y + 3) / (x + 1), y = 3x,
M coordinates of PQ midpoint: PX = (- 1 + 2) / 2 = 1 / 2, py = (6-3) / 2 = 3 / 2,
M (1 / 2,3 / 2), the slope of the vertical bisector is PQ, and the negative reciprocal of the slope of the straight line is - 1 / 3
The vertical bisector equation of PQ, (Y-3 / 2) / (x-1 / 2) = - 1 / 3,
y=-x/3+5/3,
The center coordinate of the circle is on the intersection of the vertical bisector of PQ line segment and the straight line x + 2y-4 = 0,
The intersection coordinates x = 2, y = 1,
Center coordinates C (2,1),
Circular equation: (X-2) ^ 2 + (Y-1) ^ 2 = R ^ 2,
Q-point coordinates, R ^ 2 = 25,
The circular equation is: (X-2) ^ 2 + (Y-1) ^ 2 = 25,
Distance from the center of circle C (2,1) to the straight line: D = | (k-1) * 2 + 2 * 1 + 5-3k| / √ [(k-1) ^ 2 + 4],
d=|5-k|/√（k^2-2k+5),
Square on both sides,
（5-k)^2=d^2(k^2-2k+5),
(1-d^2)k^2+2(d^2-5)k+5(5-d^2)=0,
If K has a real solution, then the discriminant △≥ 0,
d^4-5d^2≤0,
0≤d^2≤5,
0≤d≤√5,
If the center distance of the circle is the largest, the chord is the smallest,
When d = √ 5, the chord is the smallest,
Let the chord be EF,
According to Pythagorean theorem, | EF 124; / 2 = √ (R ^ 2-D ^ 2) = √ (25-5) = 2 √ 5,
∴|EF|=4√5.
Judge the position relation between the square of two circles x + the square of Y + x-2y-20 = 0 and the square of X + the square of y = 25
Please help me solve it as soon as possible
X^2+Y^2+X-2Y-20=0
(X+1\2)^2+(Y-1)^2=20+1\4+1
(X+1\2)^2+(Y-1)^2=49\4
So the radius of the circle is 7-2, and the coordinates of the center of the circle are (- 1-2,1)
X ^ 2 + y ^ 2 = 25 is a circle with (0,0) point as coordinate and border as 5
The distance between the centers of two circles is the root sign 5 divided by 2, and the radius of the former + root sign 5 divided by 2
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