On the square sheet iron with the side length of 46 cm, cut a circular sheet iron and a fan-shaped sheet iron as shown in the figure, just to make a cone model, then the bottom radius of the cone model is______ Cm

On the square sheet iron with the side length of 46 cm, cut a circular sheet iron and a fan-shaped sheet iron as shown in the figure, just to make a cone model, then the bottom radius of the cone model is______ Cm

As shown in the figure, connect AC and OE, let the bottom radius of the cone model be X. according to the equal arc length and the circumference of the bottom of the cone, the radius of the sector is 4x, then 2x + 5x = 462, and the solution is x = 102-4 (CM)

The worker should cut a complete circle and a complete fan-shaped iron sheet on the square iron sheet with a length of 40cm as shown in the figure to make it into a conical model
(1) Please help the worker to design three different cutting schemes; (2) which design scheme makes the utilization of square sheet metal the highest? The radius of the bottom circle of the cone model is obtained

(1) The schematic diagram of the design scheme is as follows. (2) ∵ ① the sector area of the figure is: 90 π × 402360 = 400 π, ② the area of the figure is: 12 π× (20) 2 + π × 102 = 300 π, ③ the sector area of the figure is: 60 π × 402360 = 800 π 3, ∵ the cutting scheme with the highest utilization rate of square sheet iron is shown in figure (1) If the circumference of the cone bottom is 14 × 2R × π = 2 π R, then r = 4R. ∵ the side length of the square is 40cm, ∵ BD = 402cm. ∵ the tangent point of O and the sector is e, the center O is on BD, ∵ R + R + 2R = 402, and the solution is r = 2002 − 8023cm

The equation of the circle with radius 5, center on Y axis and tangent to the line y = 6 is______ ．

As shown in the figure, because the radius is 5, the center of the circle is on the Y-axis and tangent to the straight line y = 6, there are two circles. The center of the upper circle is (0,11), and the center of the lower circle is (0,1), so the equation of the circle is x2 + (Y-1) 2 = 25 or x2 + (Y-11) 2 = 25

The equation of the circle with radius 5, center on Y axis and tangent to the line y = 6 is______ ．

As shown in the figure, because the radius is 5, the center of the circle is on the Y-axis and tangent to the straight line y = 6, there are two circles. The center of the upper circle is (0,11), and the center of the lower circle is (0,1), so the equation of the circle is x2 + (Y-1) 2 = 25 or x2 + (Y-11) 2 = 25

Given that a circle passing through point P (2, - 1) is tangent to a straight line X-Y = 1, and its center is on the straight line y = - 2x, the equation of circle is obtained

Let the center of the circle be (T, - 2t)
Let the equation of circle be (x-t) ^ 2 + (y + 2t) ^ 2 = R ^ 2
Then we can have the equation I: (2-T) ^ 2 + (- 1 + 2t) ^ 2 = R ^ 2
Then there is a circle tangent to a straight line, and (x-t) ^ 2 + (x-1 + 2t) ^ 2 = R ^ 2 has a unique solution
Then we can get a relation between T and R. by combining this relation with I, we can get t

The equation of a circle whose center is on the straight line 4x + y = 0 and passes through points P (4,1), q (2, - 1) is______ ．

Let the center of the circle be a (a, - 4A), then the distances from a to points P and Q are equal, and they are all equal to the radius, ｜ r = (a − 4) 2 + (− 4A − 1) 2 = (a − 2) 2 + (− 4A + 1) 2, ｜ a = - 1, so a (- 1,4), radius r = (a − 4) 2 + (− 4A − 1) 2 = 34, so the equation of the circle is (x + 1) 2 + (y-4) 2 = 34

l: Y = x, the center of the circle is (3,0), and through (4,1), find the equation of the circle
If the circle and circle 1 are symmetric with respect to the line L, a and B are circles respectively, and any two points on the circle C, the minimum value of | ab | can be obtained

Because the center of the circle is (3,0)
Let the circular equation be (x-3) ^ 2 + y ^ 2 = R ^ 2
Because by substituting the point (4,1) into the equation
(4-3)^2+1=R^2
R^2=2
So the equation of the circle is (x-3) ^ 2 + y ^ 2 = 2

The equation of the circle with points a (- 1,1), B (1,3) and the center of the circle on the x-axis

Let the standard equation of circle be: (x-a) ^ 2 + y ^ 2 = R ^ 2
Then (- 1-A) ^ 2 + 1 = R ^ 2
(1-a)^2+9=r^2
The solution is a = 2, r = root 10
So the standard equation of circle is: (X-2) ^ 2 + y ^ 2 = 10
Wuliangshou Buddha, the Buddha said that the sea of bitterness is endless, looking back is the shore!
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If I devote myself to practice, I will become a great tool in the future. I just have a book in my hand, which I want to give to the benefactor
I have a little test, please click next to the answer

What is the equation for a circle whose center is (3,2) and is tangent to the x-axis?

So the radius is two
So the equation for a circle is
(x-3)^2+(y-2)^2=4
The radius is the coordinate of Y

Through the points c (- 1,1) and D (1,3), the center of the circle is on the x-axis?

The center of the circle is on the X axis, (a, 0)
Then (x-a) ^ 2 + y ^ 2 = R ^ 2
(-1-a)^2+1^2=r^2
(1-a)^2+3^2=r^2
subtract
(-1-a)^2-(1-a)^2+1-9=0
(-1-a+1-a)(-1-a-1+a)-8=0
-2a*(-2)=8
a=2
r^2=(1-a)^2+3^2=10
(x-2)^2+y^2=10