If the equation of the circle is the square of X + the square of Y - 2x + 4y-20 = 0, find the linear equation of the diameter of the origin

If the equation of the circle is the square of X + the square of Y - 2x + 4y-20 = 0, find the linear equation of the diameter of the origin

(x-1) ^ 2 + (y + 2) ^ 2 = 25,
So the center of the circle is (1, - 2),
Therefore, the slope is k = (- 2-0) / (1-0) = - 2,
Therefore, the linear equation of the diameter passing through the origin is y = - 2x

Given that the distance between the center of circle x ^ 2 + y ^ 2-4x-4y + 4 + 2 and the line L passing through the origin is (root 3) - 1, find the equation of the line L

Let me tell you how. First find the center of a circle. This is the general equation of a circle
The origin (0.0) and (radical 3, - 1) can be written as a bivariate first-order equation
And then use the distance from the point to the straight line to get the formula. I don't list it in the book

Given that the radius of ⊙ o is 2, the distance from point P to the center of a circle OP = m, and the quadratic equation 2x ^ 2 - (2 √ 2) + M-1 = 0 of X has a real root, try to write the position of P

If the equation 2x 2 - (2 √ 2) x + M-1 = 0 has a real root, then the discriminant △ = (2 √ 2) 2-8 (m-1) ≥ 0, then m ≤ 2
Thus, the distance between point P and the center of a circle OP = m ≤ 2, that is, point P is on or inside the circle

Given that the radius r of the circle O is the root of the quadratic equation x-5x-6 = 0, and the distance between the point O and the straight line L is 8, then the position relationship between the line L and the circle O is?

R=6

If the circle represented by the equation x ^ 2 + y ^ 2 + DX + ey + F = 0 passes through the origin and the distance from the center of the circle to the two coordinate axes is equal, then D, e, f satisfy the condition that is?

When the circle passes through the origin (0,0), the equation is: F = 0
The circular equation can be simplified as: (x + D / 2) ^ 2 + (y + E / 2) ^ 2 = (e ^ 2 + f ^ 2) / 4
It is easy to know that the coordinates of the center of the circle are (- D / 2, - E / 2)
The distance to the axis is equal, so │ - D / 2 │ = │ - E / 2 │
So, d = E
To sum up: F = 0 and │ D │ = │ e │

If the square of equation x + the square of Y + DX + ey + F = 0 denotes a circle with radius 4 and C (2, - 4), then d =, e =, f =,

（x-2)²+(y+4)²=16
x²-4x+4+y²+8y+16=16
x²+y²-4x+8y+4=0
D=-4, E=8, F=4

If equation x 2 +y 2 +DX+EY+F=0 represents a circle with C (2, -4) as the center and radius of 4, then what are D+E+F=?, D, E and F equal respectively?

The sum of e4-4 is E4

If x + y + DX + ey + F = 0 is a circle with C (2, - 4) as its center and radius equal to 4, then d =, e, = f =. Who knows

Taking C (2, - 4) as the center of the circle, the equation of the circle whose radius is equal to 4 is (X-2) + (y + 4) = 16. The expansion of X + y-4x + 8y + 4 = 0 can be obtained by comparing with the original formula: D = - 4. E = 8. F = 4

It is known that the circle C: x2 + Y2 + DX + ey + 3 = 0, the circle C is symmetric about the straight line x + Y-1 = 0, the center of the circle is in the second quadrant, and the radius is Two (I) find the equation of circle C; (II) given that the straight line L which is not at the origin is tangent to the circle C, and the intercept on the x-axis and y-axis is equal, the equation of the straight line L is solved

(I) from x2 + Y2 + DX + ey + 3 = 0, we know that the coordinate of center C is (- D)
2，-E
2）
∵ the circle C is symmetric about the straight line x + Y-1 = 0
Ψ point (- D)
2，-E
2) On the line x + Y-1 = 0
That is, D + e = - 2, ① and D2 + E2 − 12
4=2②
And ∵ the center of the circle C is in the second quadrant ᙽ d > 0, e < 0
D = 2 and E = - 4 are obtained from ① and ②
The equation of circle C is: x2 + Y2 + 2x-4y + 3 = 0
Let L: x + y = a
∵ circle C: (x + 1) 2 + (Y-2) 2 = 2
The distance from the center C (- 1,2) to the tangent line is equal to the radius
2，
That is − 1 + 2 − a
2|=
2, ν a = - 1 or a = 3
The tangent equation x + y = - 1 or x + y = 3

In the known triangle ABC, ∠ C = 90 ° ad is the bisector of angle, CD = 6cm, BD = 10cm

Make a vertical line to ad through point D, and the vertical foot is e
Because AD is the angle bisector
So ∠ CAD = ∠ bad
Because ∠ C = ∠ AED = 90 ° ad = Da
So RT △ ACD ≌ RT △ ade
AE = 6, so CD = 6
From Pythagorean theorem
Be square = BD square - ed square = 64
BE＝8
Let AC = AE = x, then AB = x + 8
x2＋16 2＝（x＋8）2
The solution is x = 12
So AC = AE = 12
PS: X2, 162, (x + 8) 2 all denote the square of a number!