(2006 Yantai) known: on the one variable quadratic equation X2 - (R + R) x + 1 4D2 = 0 has no real root, where R and R are the radii of ⊙ O1, ⊙ O2 respectively, and D is the center distance between the two circles, then the position relationship of ⊙ O1, ⊙ O2 is () A. Exotropism B. Tangency C. Intersection D. Contains

(2006 Yantai) known: on the one variable quadratic equation X2 - (R + R) x + 1 4D2 = 0 has no real root, where R and R are the radii of ⊙ O1, ⊙ O2 respectively, and D is the center distance between the two circles, then the position relationship of ⊙ O1, ⊙ O2 is () A. Exotropism B. Tangency C. Intersection D. Contains

According to the meaning of the question, if the equation has no real root, we can get (R + R) 2-d2 < 0,
Then: (R + R + D) (R + R-D) < 0,
Because R + R + d > 0, R + R-D < 0,
That is, d > R + R,
Then, the two circles are separated
Therefore, a

If the diameter of the circle O is 2, the distance from the center of the circle O to the straight line L is m. if there is no real root in the quadratic equation MX? - 2 of X, there is no real root, then the position relationship between circle O and line L is () A. Intersect B. separate C. tangent D. intersect or tangent

(-2√2)²-4m×2

(2006 Yantai) known: on the one variable quadratic equation X2 - (R + R) x + 1 4D2 = 0 has no real root, where R and R are the radii of ⊙ O1, ⊙ O2 respectively, and D is the center distance between the two circles, then the position relationship of ⊙ O1, ⊙ O2 is () A. Exotropism B. Tangency C. Intersection D. Contains

According to the meaning of the question, if the equation has no real root, we can get (R + R) 2-d2 < 0,
Then: (R + R + D) (R + R-D) < 0,
Because R + R + d > 0, R + R-D < 0,
That is, d > R + R,
Then, the two circles are separated
Therefore, a

Given that the radius of the circle O is r, the distance between the line AB and the center O of the circle is D, and the equation x? - 2 √ DX + r = 0 has real roots, then the position relationship between the line AB and the circle O is?

If there is a real root, B ^ 2-4ac > = 0
The substitution is 4d-4r > = 0
d>=R
That is, the relationship between a line and a circle is tangent or apart

Given that the absolute value of OA vector = 1, the absolute value of OB vector = root 3, OA dot multiplication ob = 0, point C is in the angle AOB, and the angle AOC = 30 °, let OC vector = MOA + nob to find M / n The following answer is very difficult to understand, can you explain it in your own words

Answer: M / N = 3 / x0d please refer to the figure below for details

Let the convex quadrilateral ABCD circumscribed to the circle O, and ab = 2, BC = 3, CD = 4, then OA * OC + ob * od = (6 * radical 2), why? Ao = root 3, OC = root 6, OA * OC + ob * od = 2 * Radix 3 * Radix 6 = 6 * Radix 2 How did this come about?

Since the sum of the opposite sides of a circle circumscribed quadrilateral is equal, it is easy to know that Da = 3
So the quadrilateral ABCD is an isosceles quadrilateral, its height is the diameter of circle O, so it is easy to calculate the radius of circle O as the root 2. Then the length of Ao, OC and other segments can be easily calculated

The eccentricity of the ellipse is root 6 / 3, and the distance from one end of the minor axis to the right focus is root 3 The eccentricity of ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is root 6 / 3, the distance from one end point of minor axis to right focus is root 3, (1) solve elliptic equation (2) let line L and ellipse C intersect at two points AB, the distance from coordinate origin o to straight line L is root 3 / 2, and the maximum area of triangle AOB is obtained

(1) the distance from one end of the minor axis to the right focus is root 3, that is, a = 3
From E = C / a = root 6 / 3, C = root 6
a^2=9,c^2=6,b^2=a^2-b^2=3,
So the equation is x ^ 2 / 9 + y ^ 2 / 3 = 1

It is known that the eccentricity e of the ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is known to be the root of 3, and the distance from one end point of the minor axis to the right focus is root 3 Find (1): the equation of elliptic C (2) : let the line L and ellipse C intersect at two points a and B, and the distance from coordinate origin o to line L is the root of 2 / 3, and the maximum area of triangle ABC is obtained

(1):
e²=c²/a²=2/c
b²+c²=3
b²+a²-b²=3
a²=3,b²=1
The elliptic equation is x 2 / 3 + y 2 = 1

Given that the ellipse C: x square / a square + y square / b square = 1, the eccentricity is 5 / 3, and the distance from one end point of the minor axis to the right focus is 3, the equation of ellipse C is obtained,

One end point of minor axis B (0, b) right focus F2 (C, 0)
|BF2|=√(b^2+c^2)=a=3
e=c/a=√5/3 c=√5
b^2=a^2-c^2=4
The equation of elliptic C
x^2/9+y^2/4=1

It is known that the eccentricity of the ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is 2 / 2 root 3, and the distance from one end of the minor axis to the right focus is 2 If P is a moving point on the ellipse, F1 and F2 are the left and right focal points of the ellipse respectively. Find the maximum and minimum values of vector Pf1 * vector PF2

The distance from one end of the minor axis to the right focus is a a = 2 C = √ 3 / 2, B = 1 ellipse x ^ 2 / 4 + y ^ 2 = 1 F1 (- √ 3,0) F2 (√ 3,0) P (m, n) vector Pf1 = (- M - √ 3, - n) vector Pf1 = (- M + √ 3, - n) vector Pf1 * vector PF2 = m ^ 2 + n ^ 2-3m ^ 2 / 4