As shown in the figure, O is a point on the diagonal of the square ABCD. The circle with o as the center and OA length as the radius is tangent to m and AB, and ad intersects EF respectively (2) If the side length of the square is 1, find the o radius of the circle

It doesn't matter if the diagonal is AC
Connect OM, because the circle O and BC are tangent to m, so OM is perpendicular to BC. Because they are radius, OM = OA;
Let OA = x, then om = x, because AB = 1, so diagonal = root 2, OC = root 2-x, because angle ACB = 45 degrees, so root 2 times x = root 2-x, x = 2-radical 2
So radius = 2 - radical 2

As shown in the figure, O is a point on the diagonal of square ABCD, and ⊙ o with o as the center and OA length as the radius is tangent to point M (1) Verification: CD is tangent to ⊙ o (2) If the side length of the square ABCD is 1, find the radius of ⊙ o

It is proved that: (1) connect om with O to make on ⊥ CD in n;
⊙ o is tangent to BC,
∴OM⊥BC，
∵ the quadrilateral ABCD is a square,
 AC bisection ∠ BCD,
∴OM=ON，
⊙ CD is tangent to ⊙ o
（2） ∵ the quadrilateral ABCD is a square,
∴AB=CD=1，∠B=90°，∠ACD=45°，
∴AC=
2，∠MOC=∠MCO=45°，
∴MC=OM=OA，
∴OC=
OM2+MC2＝
2ON＝
2OA；
And ∵ AC = OA + OC,
∴OA+
2OA=
2，
∴OA=2-
2．

As shown in the figure, O is a point on the diagonal AC of square ABCD, and ⊙ o with o as the center and OA length as the radius is tangent to point M Verification: CD is tangent to ⊙ o

It is proved that when OM is connected, point O is on ⊥ CD is at point n,
⊙ O and BC are tangent to point M,
∴OM⊥BC，
And ∵ on ⊥ CD, O is a point on the diagonal AC of square ABCD,
∴OM=ON，
⊙ CD is tangent to ⊙ o

As shown in the figure, O is a point on the diagonal AC of square ABCD, and ⊙ o with o as the center and OA length as the radius is tangent to point M Proof: circle O is tangent to CD

Draw the plumb line from point O to CD, and the foot of the plumb is point n, that is, it intersects with point n on CD;
In triangle OCM and triangle OCN, because angle com = angle con = 90 degrees, angle ACB = angle ACD, OC = OC, so triangle OCM and triangle OCN are congruent;
So ON=OM= radius of circle, and because ON is perpendicular to CD, circle O is tangent to CD, and the tangent point is N

It is known that the center of the circle O is at the origin of the rectangular coordinate system, the radius is 1, and the point P is a moving point on the circle O (not on the coordinate axis), It is known that the center of circle O is at the origin of Cartesian coordinate system, the radius is 1, and point P is a moving point on circle O (not on the coordinate axis). Let the tangent line of circle O passing through point P intersect with points a and B respectively with X and Y axes. When point P moves, there exists___ There are three triangles AOB whose area is exactly 4

Looks like eight
2 quadrants each
These two are about the quadrant angle bisector axis symmetry~
To be honest, it's better to ask a teacher than a netizen

As shown in the figure, make a circle with the origin o of the rectangular coordinate system as the center of the circle, a is a point on the X axis, AB tangent circle O is at point B, if AB = 12, ad = 8, calculate the coordinates of B Sorry, I can't draw Please fill in the figure below [b is the intersection of tangent and circle

What is d? Because ad = 8, is it Ao?

In Cartesian coordinate system, the radius of circle O is 1. Judge the position relationship between the straight line y = - x + root 2 and circle O, and explain the reason

In Cartesian coordinate system, the radius of circle O is 1. Judge the position relationship between the straight line y = - x + root 2 and circle O, and explain the reason
It is easy to find the length of the line y = - x + root 2 by passing o
So the line y = - x + root 2 is tangent to the circle o

There is a point a (3,4) in the plane rectangular coordinate system. A is the center of the circle and 5 is the radius to draw a circle. In the same coordinate system, the position relationship between the straight line y = - X and ⊙ A is () A. Separation B. Tangency C. Intersection D. All of the above are possible

As shown in the figure,
∵A（3，4），∴AO=5，
∵ the distance from point a to line y = - x is that the length of AB is less than the radius of circle R, that is, AB < Ao,
⊙ the position relationship between the line y = - X and ⊙ A is intersection,
Therefore, C

There is a point a (3,4) in the plane rectangular coordinate system. A is the center of the circle and 5 is the radius to draw a circle. In the same coordinate system, the position relationship between the straight line y = - X and ⊙ A is () A. Separation B. Tangency C. Intersection D. All of the above are possible

In the rectangular coordinate system, take the points a (0, 3), B (4, 0) as the center of the circle, and 6 and 1 as the radius to make ⊙ A and ⊙ B, respectively______ .

∵ points a (0, 3), B (4, 0),
∴AB=
32+42=5，
The radii of ⊙ A and ⊙ B are 6 and 1 respectively,
The radius difference is: 6-1 = 5,
The relationship between the two circles is: inscribed
So the answer is: introversion

As shown in the figure, O is a point on the diagonal of the square ABCD. The circle with o as the center and OA length as the radius is tangent to m and AB, and ad intersects EF respectively (2) If the side length of the square is 1, find the o radius of the circle