It is known that the straight line L is tangent to the circle C: x square + y square + 2x-4y + 4 = 0, and the distance from the origin o to L is 1, and the equation of the straight line L is obtained

It is known that the straight line L is tangent to the circle C: x square + y square + 2x-4y + 4 = 0, and the distance from the origin o to L is 1, and the equation of the straight line L is obtained

(x+1)²+(y-2)²=1
Drawing the graph, we can get the line y = 1
In addition, there is another formula, which is based on the distance between a point and a straight line, which can be quickly obtained

If the distance from the center of the circle x-y-2x-4y = 0 to the straight line passing through the origin is 1, then the equation of the line is( As the title

Square of circle x + square of Y - 2x-4y = 0
(x-1)^2+(y-2)^2=5
Center (1,2)
The linear equation y = KX passing through the origin
|k-2|/√(1+k^2)=1
(k-2)^2=1+k^2
-4k+4=1
k=3/4
This linear equation is y = 3x / 4

Given the circle C: x2 + Y2 + 2x + 4Y + 1 = 0, then the linear equation passing through the center of the circle C and having the largest distance from the origin is______ .

The equation of circle C can be changed into (x + 1) 2 + (y + 2) 2 = 4, so the coordinates of the center of the circle are (- 1, - 2) the slope of the line connecting the center of the circle and the origin is − 2 − 0 − 1 − 0 = 2, and the slope of the line with the largest distance from the origin is − 12, and the line passes through the center (- 1, - 2), so the equation is Y - (- 2) = −

It is known that the radius of circle a is 1, the distance between point P and point O is r, and the quadratic equation x 2 - 2x + r = 0

Because the equation has real roots, so △ = 4-4r ≥ 0, R ≤ 1,
The radius of the circle is 1, which means that the distance from point P to the center of the circle is less than or equal to the radius, so point P is in or on the circle

If the circle C with radius r, x ^ 2 + y ^ 2 + DX + ey + F = 0, the distance between the center C of circle C and the straight line L: DX + ey + F = 0 is D, where d ^ 2 + e ^ 2 = f ^ 2 and F > 0 1. Find the range of F 2. Verification: D ^ 2-r ^ 2 is a constant value

(1) (2) substituting the center of circle (- D / 2, - E / 2) into the formula of the distance between the point and the line, d = | - D? 2-e? 2 / 2 + F | / √ (D | + e ﹤)

It is known that the radius of the circle O is 10, the distance from the center of the circle O to the straight line L is od = 6. There are three points a, B and V on the line L, ad = 6, BD = 8, CD = 5 times the root sign 3 Question: what is the position relationship between point a, B and C and circle O? Question: what is the relationship between a, B, C (the number above is wrong, not V is c) and the position of circle O?

∵ OA = 6 times root number, OB = 10, OC = root number 111, OA = 6 times root number 2 is less than R, OB = R, OC is greater than R
⊙ o, B ⊙ o, C ⊙ o
I made it myself. I don't know if it's right!

It is known that the radius of circle O is 10, the distance od from the center of circle O to the straight line a is 6 cm, there are three ABC points on the line a, and there are ad = 10 cm, BD = 8 cm, CD = 6 cm, respectively Point out the position relation of point a, B, C and circle o

V. the radius of ⊙ o is 10 cm, the distance from center O to line L is od = 6 cm, there are three points a, B and C on the line L, and there are ad = 10 cm, BD = 8 cm, CD = 6 cm. The location relations of points a, B, C and ⊙ o are pointed out respectively
[process] let students draw a graph, combine number and shape, according to Pythagorean theorem, we can get OA = cm, OB = 10cm, OC = cm respectively, and then compare OA, ob, OC and radius
[results] point a is outside ⊙ o, point B is on ⊙ o, and point C is in ⊙ o

If the straight line y = (radical 3 / 3) x + radical 2, and the circle (x-radical 3) ^ 2 + (Y-1) ^ 2 = 3 with the center of the circle D intersect at a and B, then the sum of inclination angles of AD and BD is The answer is that four thirds of PI is urgent

Is ad and straight line?

It is known that the radius of circle O is 5cm, and the length of OD from point O to line L is 4cm. If points a, B and C are on line L, and ad = double root 2cm, If BD = double root 3cm, CD = 3cm, then point a is at, point B is at, and point C is at

Inside, outside, round

It is known that circle C passes through points a (1,4) and B (3, - 2), and the distance between the center of circle C and the straight line AB is Find the equation of circle C

Method I: let C (a, b) be the center of the circle, and the radius is r
The midpoint of the easy to see line AB is m (2,1) (2 points)
∵CM⊥AB,kAB=-2-4
3-1=-3
∴kCM=b-1
a-2=1
3 is: 3B = a + 1 (5 points)
And ∵ cm|=
10∴(a-2)2+(b-1)2=10②… (8 points)
We can get
a=-1
B = 0 or
A=5
B=2
That is C (- 1,0) or C (5,2) (10 points)
∴r2=|CA|2=20
So the equation of the circle is: (x + 1) 2 + y2 = 20 or (X-5) 2 + (Y-2) 2 = 20 (12 points)
Method II: ∵ a (1,4), B (3, - 2)
The equation of line AB is: 3x + Y-7 = 0 (2 points)
∵ the midpoint of line AB is m (2,1)
The center of the circle C falls on the perpendicular line of the line AB: x-3y + 1 = 0 (4 points)
Let C (3b-1, b) (5 points)
∴|3(3b-1)+b-7|
32+12=
10… (8 points)
The solution is b = 0 or B = 2
That is C (- 1,0) or C (5,2) (10 points) ν R2 = | Ca | 2 = 20
So the equation of the circle is: (x + 1) 2 + y2 = 20 or (X-5) 2 + (Y-2) 2 = 20 (12 points)