The radius of a circle is 5cm, the degree of sector is 60, and what is the sector area

The radius of a circle is 5cm, the degree of sector is 60, and what is the sector area

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3.14×5×5×60/360
=13.086 square centimeter
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For a 90 degree sector, the diameter of the circle is 2.5cm. What is the area of the sector?
The π is 3.14

2.5 △ 2 = 1.25 cm
3.14×1.25×1.25×90/360
=1.2265625 square centimeter

Cut a sector with a central angle of a on the circular iron sheet with radius r to make the remaining part form a cone. What is the maximum volume of the cone when a is the value?

Let the radius of the bottom of the cone be r, r = x * r, then x = 1-A / 2 π can be calculated
The volume V of cone is π R & sup2 * (R & sup2-r & sup2) & sup0.5
v=πx²*R²*（R²-x²*R²）&sup0.5
v=πR³*（x&sup4-x&sup6）&sup0.5
V '= 4x & sup3-6 & sup5 is obtained by deriving the variable X & sup4-x & sup6
Let V '= 0 get x = 6 & sup0.5/3
According to x = 1-A / 2 π, a can be obtained

Finding the coordinates of the center of a circle by the equation of a known circle
Find the center coordinate of x ^ 2 + y ^ 2 + 2x = 0 as?

x^2+y^2+2x=0
Formula: (x + 1) ^ 2 + y ^ 2 = 1
Center (- 1,0)

The center of the circle (2, 3), through the coordinate origin to find the equation of the circle

r²=(2-0)²+(3-0)²=13
So (X-2) ² + (Y-3) ² = 13

The equation of a circle whose center is on the y-axis, radius is 1 and passes through point (1,2) is ()
A. x2+（y-2）2=1B. x2+（y+2）2=1C. （x-1）2+（y-3）2=1D. x2+（y-3）2=1

Solution 1 (direct method): let the coordinates of the center of a circle be (0, b), then we know (o − 1) 2 + (B − 2) = 1 from the meaning of the title, and we get b = 2, so the equation of a circle is x2 + (Y-2) 2 = 1. So we choose A. solution 2 (combination of number and shape): from the distance from the point (1, 2) to the center of a circle is 1, we can easily know that the center of a circle is (0, 2), so the equation of a circle is X

The equation of a circle whose center is on the y-axis, radius is 1 and passes through point (1,2) is ()
A. x2+（y-2）2=1B. x2+（y+2）2=1C. （x-1）2+（y-3）2=1D. x2+（y-3）2=1

Solution 1 (direct method): let the coordinates of the center of a circle be (0, b), then we know (o − 1) 2 + (B − 2) = 1 from the meaning of the title, and we get b = 2, so the equation of a circle is x2 + (Y-2) 2 = 1. So we choose A. solution 2 (combination of number and shape): from the distance from the point (1, 2) to the center of a circle is 1, we can easily know that the center of a circle is (0, 2), so the equation of a circle is X

The equation of a circle whose center is on the y-axis, radius is 1 and passes through point (1,2) is ()
A. x2+（y-2）2=1B. x2+（y+2）2=1C. （x-1）2+（y-3）2=1D. x2+（y-3）2=1

Solution 1 (direct method): let the coordinates of the center of a circle be (0, b), then we know (o − 1) 2 + (B − 2) = 1 from the meaning of the title, and we get b = 2, so the equation of a circle is x2 + (Y-2) 2 = 1. So we choose A. solution 2 (combination of number and shape): from the distance from the point (1, 2) to the center of a circle is 1, we can easily know that the center of a circle is (0, 2), so the equation of a circle is X

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