Cut a circle and a sector from the square sheet iron to form a cone (as shown in the figure). If the radius of the circle is R and the radius of the sector is r, then the radius of the circle accounts for half of the radius of the sector______ %．

Cut a circle and a sector from the square sheet iron to form a cone (as shown in the figure). If the radius of the circle is R and the radius of the sector is r, then the radius of the circle accounts for half of the radius of the sector______ %．

Because the arc length of the sector is equal to the circumference of the bottom of the cone, so 14 × 2 π r = 2 π R, simplified to: r = 4R, r = 14R, that is, the radius of the circle accounts for 14 = 25% of the sector radius; answer: the radius of the circle accounts for 25% of the sector radius; so the answer is: 25

If the arc length of a sector with a central angle of 60 ° is 2 π, its inscribed circle radius is ()
A. 2B. 3C. 1D. 32

Let the radius of sector and inscribed circle be r, R respectively. From 2 π = π 3R, the solution is r = 6. ∵ 3R = r = 6, ∵ r = 2

The equation of the circle passing through two points (3,5), (- 3,7) and the center of the circle on the x-axis is______ ．

Let C (a, 0) be the center of the circle. From the distance formula of two points, we can get | Ca | = (3 − a) 2 + 52, | CB | = (− 3 − a) 2 + 72 ∵ two points a (3, 5), B (- 3, 7) on the circle, we can get (3 − a) 2 + 52 = (− 3 − a) 2 + 72, we can get the center C (- 2, 0), radius r = 50 = 52, so we can get the circle

If the center of a circle is on the x-axis, the radius is equal to 5 and passes through (6,3), the equation of the circle is obtained

Center (a, 0)
r²=25
So (a-6) & sup2; + (0-3) & sup2; = 25
a²-12a+36+9=25
(a-2)(a-10)=0
therefore
(X-2) & sup2; + Y & sup2; = 25 and (X-10) & sup2; + Y & sup2; = 25

The equation of finding a circle on the y-axis through the center of a (5,2) and B (- 3,0)

According to the midpoint coordinate formula, the midpoint of AB is (5 + (- 3) / 2, (2 + 0) / 2), i.e. (1,1)
The slope of line AB is k = (2-0) / (5 - (- 3)) = 1 / 4
So the slope of the vertical bisector of AB is - 1 / k = - 4
So the equation of the vertical bisector of AB written in the oblique form of a point has the following characteristics
Y-1 = - 4 (x-1), that is y = - 4x + 5
The intersection of the line and the y-axis is the center coordinate of the circle, which is (0,5)
Therefore, according to the formula of distance between two points, the radius of circle is √ (5-0) & sup2; + (2-5) & sup2; = √ 34
According to the standard equation of a circle, there are
(x-0)²+(y-5)²=34
That is, X & sup2; + (Y-5) & sup2; = 34 is the equation of the circle

Find the equation of circle passing through point B (- 2, - 4) and intersecting with line x + 3y-26 = 0 and point a (8,6)

There are countless circles passing through point B (- 2, - 4) and intersecting with line x + 3y-26 = 0 and point a (8,6),

Through a (- 1,5) B (5,5) C (6, - 2), find the equation of circle
THANK YOU!1

What is the equation of a circle
=_ =If it's the center coordinates, I can tell you
(3,2)

The equation of a circle passing through three points a (- 1,5), B (5,5), C (6, - 2) is______ ．

Let the equation of the circle passing through three points a (- 1,5), B (5,5), C (6, - 2) be x2 + Y2 + DX + ey + F = 0. By substituting the coordinates of these three points into the equation, we can get 1 + 25 − D + 5E + F = 025 + 25 + 5D + 5E + F = 036 + 4 + 6D − 2E + F = 0

The equation of circle passing through points a (- 1,5), B (5,5), C (6, - 2)

Through the point a (- 1,5), B (5,5), (- 1 + 5) / 2 = 2, the center of the circle is on x = 2
Let the circular equation be (X-2) ^ 2 + (y-b) ^ 2 = R ^ 2
Substituting (5,5) and (6, - 2) to get
9+(5-b)^2=r^2
16+(-2-b)^2=r^2
The solution is b = 1
We can get r = 5
So the equation of circle is (X-2) ^ 2 + (Y-1) ^ 2 = 25

Find the equation of a circle passing through three points a (- 1,5), B (5,5), C (6, - 2)?

If passing through points a (- 1,5), B (5,5), (- 1 + 5) / 2 = 2, then the center of the circle is on x = 2, let the circular equation be (X-2) ^ 2 + (y-b) ^ 2 = R ^ 2, substitute (5,5) and (6, - 2) to get 9 + (5-b) ^ 2 = R ^ 216 + (- 2-B) ^ 2 = R ^ 29 + (5-b) ^ 2 = 16 + (- 2-B) ^ 2, and get b = 1 to get r = 5, so the circular equation is (X-2) ^ 2 + (Y-1) ^ 2 = 25