If the radii of two tangent circles are two of the equations x2-3x + 2 = 0, then the value of the center distance d between the two circles is______ ．

If the radii of two tangent circles are two of the equations x2-3x + 2 = 0, then the value of the center distance d between the two circles is______ ．

∵ the radii of the two circles are two of the equation x2-3x + 2 = 0, ∵ R1 = X1 = 1, R2 = x2 = 2, ∵ the two circles are tangent, ∵ the center distance of the circle is 2-1 or 2 + 1, so the answer is: 1 or 3

The expression of R is given as 1 / R 1 = 1 / R 1 + 1 / R 2 (R 1 + R 2 ≠ 0)

General division
1/R=(R1+R2)/R1R2
R=R1R2/(R1+R2)

It is known that the radius of circle O1 and circle O2 is R1, R2, R1 = 2, R2 = 3, and the center distance D satisfies 1 = or

Yes, when D is equal to 2 or 5, it is a common point, and when D is between 1 or 5, there are two common points!

Let the radius of circle O1 and circle O2 be r1r2, and use equality or inequality to describe the position relationship between circle O1 and circle O2

Let the center distance of the circle be l R1 > R2
R1 + R2 > L phase separation
R1 + R2 = l circumscribed
R1-R2 > L intersection
R1-R2 = l inscribed
R1-R2

Given the center distance D of circle O1, circle O2, circle O2 = 2, the radius R2 of circle O2 = 3, and the value of radius R1 of circle O1, the two circles are respectively inscribed, containing?

When R1 is 1 or 4, it is inscribed, and when R1 is greater than 0 but less than 1 or greater than 5, it is included

In the triangle ABC, ∠ C = 90 °, ab = 3, BC = 2, taking point a as the center of the circle and 2 as the radius as the circle, then the position relationship between point C and center a is
A. Point C is on Center A. B. point C is outside center A. C. point C is in center A. D. cannot be determined

B
Make a triangle
According to BC ^ + AC ^ = ab^
ac^=3^-2^
ac^=5
AC = root 5
Because the root 5 is greater than the radius 2 of the circle
So point C is outside center a

If three sides a, B and C of a triangle satisfy (a-b) (B-C) (C-A) = 0, then its shape is ()
A. Right triangle B. isosceles triangle C. isosceles right triangle D. equilateral triangle

∵ (a-b) (B-C) (C-A) = 0, ∵ A-B = 0 or B-C = 0 or C-A = 0, ∵ a = B or B = C or C = a, ∵ ABC is isosceles triangle, so B

In the triangle ABC, the angle a = 2, the angle B = 3, the angle C____ triangle.
I figured it out, but angle a is not equal to twice of angle 2. What's the matter?

Let C = X
x+3x/2+3x=180
x=360/11
Angle c = 360 / 11, B = 540 / 11, a = 1080 / 11
Maximum angle a > 90
Obtuse triangle

It is known that in the triangle ABC, the angle c is equal to 90 degrees, the lengths of the three sides are a, B, C, and R are the radii of the inscribed circle. The proof of R = 1 / 2 (a + B-C) is solved by the method of the third elementary school
It is known that in the triangle ABC, the angle c is equal to 90 degrees, the length of three sides is a, B, C, and R is the radius of the inscribed circle
The third part of AB is solved by the method of elementary three

You draw the picture and then make the vertical lines on each side through the center of the circle. After the vertical lines are finished, connect the corners and the center of the circle (also the bisector of the corners)
You will find that C = A-R + B-R = a + b-2r
Then a + B-C = a + B-A + R-B + r = 2R, so r = 1 / 2 (a + B-C)

The root of equation x ^ 2 + 2006-2008 = 0, try to find the value of (m ^ 2 + 2006-2007) (n ^ 2 + 2006 + 2007)

Because m and N are the roots of the equation x ^ 2 + 2006x-2008 = 0
Therefore, m ^ 2 + 2006m-2008 = 0 can be changed into m ^ 2 + 2006m-2007 = 1
N ^ 2 + 2006n-2007 = 0 is n ^ 2 + 2006n + 2007 = 4015
So (m ^ 2 + 2006m-2007) (n ^ 2 + 2006n + 2007) = 1 * 4015 = 4015