Given that ⊙ O1 and ⊙ O2 are circumscribed at point a, the radius of ⊙ O1 is r = 2, and the radius of ⊙ O2 is r = 1, then the circle with radius 4 has () A. 2 B. Four C. Five D. 6

There are six,
And two circles are inscribed,
There are two that are circumscribed with one circle and inscribed with another,
And two circles are circumscribed,
2+2+2=6，
Therefore, D

Given that the radius of circle O1 and circle O2 are 2 and 5 respectively, and the distance between the centers of circles O1O2 = 3, then the position relationship between the two circles is

The inscribed 5-3 = 3 is equal to the center distance of the circle

If O1O2 = 5cm, the radius of circle O1 is 7cm, then the radius of circle O2 is_____ .

2 or 12

If the radius of circle O1 is 3cm and O1O2 = 1, what is the radius of circle O2? Circle and the position of the circle in that section

The distance between the centers of an inscribed circle is equal to the difference in radius
∴|R-3|=1,
R = 4 or 2

It is known that if the circle O1 is inscribed with circle O2, O1O2 = 5cm, and the radius of circle O1 is 7cm, then the radius of circle O2 is Or__ .

2 or 12

It is known that the radii of circle O1 and circle O2 are R and R respectively (r > R), and the distance between centers of circles is d. if two circles intersect, try to judge the root of equation x ^ 2-2 (D-R) x + R ^ 2 = 0

1. When O1 and O2 are circumscribed, R + r = D,
The original equation △ = 4 (D-R) - 4R? 2 = 4R? 2 = 0, that is, there is only one root;
When R, O 2, r = 1
The original equation △ = 4 (D-R) 2 - 4R 2 = 4 (- R) 2 - 4R 2 = 0, that is, there is only one root;
3. When O1 and O2 only intersect, R-R < d < R + R, then - R < D-R < R, | D-R | R, so | D-R | R |,
The original equation △ = 4 (D-R) 2 - 4R? 2 ＜ 4R? 2 = 0, i.e. rootless;
In conclusion, when o 1 and O 2 are inscribed or circumscribed, the equation has only one root; when o 1 and O 2 only intersect, the equation has no solution

The radius of O1 and O2 are known to be r, and the distance between the centers of R (r > R) is D, and the two circles intersect, we can judge the case of the root of X? 2 (D-R) x + R

∵ two circles intersect,
∴R－r＜d＜R＋r
ν D - (R-R) > 0 and D - (R + R) < 0
From the discriminant formula of the root of the square, we can get the following conclusion
∴Δ=[－2﹙d－R﹚]²－4r²
=4﹙d－R＋r﹚﹙d－R－r﹚
=4[d－﹙R－r﹚][d－﹙R＋r﹚]＜0
The equation has no real roots

Given that the radius of O1 and O2 are R, R (r > R), the center distance of the circle is D, and the two circles intersect, we can judge the situation of the root of the quadratic equation of one variable about X Given that the radius of O1 and O2 are R, R (r > R), the distance between the centers of the circle is D, and the two circles intersect, we can judge the case of the univariate quadratic equation x ^ - 2 (D-R) + R ^ = 0

Discriminant = 4 * (D-R) ^ 2 - 4 * R ^ 2
Two circles intersect, so R-R < d < R + R
When d = R, the minimum value of the equation is - 4 * R ^ 2

The radius of circle O1 and circle O2 are R and R, respectively. The distance between centers of circles is D. the two circles are separated. Point P moves on circle O1 and point Q moves on circle O2. The maximum and minimum of PQ are obtained

Using the combination of number and shape
Drawing can tell
When P and Q move to the line of the center of the circle, and P and Q are at the near end of the circle
Minimum PQ distance = d-r-r
When moving to the far end,
PQ distance max = D + R + R

The radius of circle O1 and circle O2 are 1 and 2 respectively, and the absolute value of O1O2 is 4. The moving circle is tangent to circle O1 and circumscribed to O2, which is the center locus of the moving circle I want a hyperbolic answer

It should be a branch of hyperbola. If the radius of the moving circle r O3 o3o1 is D1 and o3o2 is D2, then d2-d1 = (R + 2) - (R-1) = 3

Given that ⊙ O1 and ⊙ O2 are circumscribed at point a, the radius of ⊙ O1 is r = 2, and the radius of ⊙ O2 is r = 1, then the circle with radius 4 has () A. 2 B. Four C. Five D. 6