The arc length and area of a sector are both 5, and the arc number of the central angle of the sector is 5______ ．

Let the arc number of the central angle of the sector be α and the radius be r. ∵ the arc length and area of a sector are 5, ∵ 5 = α R, 5 = 12 α R2, and the solution is α = 52

It is known that the length of an arc is 6.28 cm and the radius of the circle is 4 cm

4×2×3.14=18.84cm
18.84÷6.28=

A circle passes through point P (2,1), the center of the circle is on the straight line x + 2y-1 = 0, and the radius is 3
Write the process

Let the center of the circle be (1-2a, a) the circle equation (x + 2a-1) ^ 2 + (Y-A) ^ 2 = 9, and the circle pass through the point P (2,1) to substitute the point P into the circle equation (1 + 2a) ^ 2 + (1-A) ^ 2 = 9, and the solution is A1 = - 1, A2 = 7 / 5, so the circle equation (x-3) ^ 2 + (y + 1) ^ 2 = 9 or (x + 12 / 5) ^ 2 + (Y-7 / 5) ^ 2

Find the equation of the circle whose center is on the straight line x-3y = 0 and passes through the origin and the point (- 4.2)

Let the circular equation be (x-a) ^ 2 + (y-b) ^ 2 = R ^ 2
From the meaning of the title
A-3b = 0, so a = 3B
Again
（0-a）^2+(0-b)^2=r^2
（-4-a）^2+(2-b)^2=r^2
From the above three formulas, we can get a = 3, B = 1, r = root 10
So the equation is: (x-3) ^ 2 + (Y-1) ^ 2 = 10

Through the coordinate origin and point P (1,1), and the center of the circle is on the line 2x + 3Y = 0, find the circle

Because the center of the circle is on the straight line 2x + 3Y = 0, let the center coordinate be (- 3T, 2t). The distance from the point to the origin of the coordinate is the same as the distance to the point P. so we get the equation: (- 3T) square + (2T) square = (- 3t-1) square + (2t-1) square solution: T = - 1, so the center coordinate is: (3, - 2) the radius of the circle is

The equation of a circle is obtained according to the following conditions: it passes through the coordinate origin and point P (1,1), and the center of the circle is on the straight line 2x + 3Y + 1 = 0;

Center (4, - 3) r = 5

Solve the equation of the circle with P (5, - 3), q (0,6) and the center of the circle on the line L: 2x-3y-6 = 0

Let the center coordinates of the circle be o (a, b), then OP = OQ
(a-5)²+(b+3)²=a²+(b-6)²
→ 9b=5a+1 …… ①
O on the line L: 2x-3y-6 = 0
→ 2a-3b-6=0 …… ②
A = 19, B = 32 / 3
r²=a²+(b-6)²=3445/9
That is to say, the circular equation is (X-19) & sup2; + (y-32 / 3) & sup2; = 3445 / 9

Solve the equation of a circle with two points P (5, - 3), q (0,6) and the center of the circle on 2x-3y-6 = 0
The answer is X & # 178; + Y & # 178; - 38x - (64 / 3) y + 92 = 0,

The vertical line of PQ passes through the center of circle
The coordinates of PQ midpoint are (5 / 2,3 / 2) and the slope is - 9 / 5
The vertical equation is y = 5 / 9 (X-5 / 2) + 3 / 2, that is 9y-5x-1 = 0
The intersection point of the line and 2x-3y-6 = 0 is a circle point
The point of intersection is O (19,32 / 3)
The radius square is Qo ^ 2 = 19 ^ 2 + (32 / 3-6) ^ 2 = 19 ^ 2 + (32 / 3) ^ 2 + 36-128 = 19 ^ 2 + (32 / 3) ^ 2-92
The equation of circle is (X-19) ^ 2 + (y-32 / 3) ^ 2 = 19 ^ 2 + (32 / 3) ^ 2-92
That is, X & # 178; + Y & # 178; - 38x - (64 / 3) y + 92 = 0

Given that the circle passes through the point (1,2), the center of the circle is on the x-axis and tangent to the line 3x + 4Y-2 = 0, the equation of the circle is obtained
I made Delta

Not really
But the topic is not difficult, draw a straight line, then the distance from a point to the straight line on the X axis is equal to the distance to the point (1,2)
And the title is right, there must be a solution, you can calculate it
Oh, I'm sorry. Just now, it's a misunderstanding

The equation of a circle whose center is on the positive half axis of X axis, radius is 3 and tangent to the line 3x + 4Y + 4 = 0 is___ ．

According to the meaning of the question, let the coordinates of the center of the circle be (a, 0) (A & gt; 0), the radius r = 3, the equation of the circle is (x-a) 2 + y2 = 3, the line 3x + 4Y + 4 = 0 is tangent to the circle, and the distance from the center of the circle to the straight line d = | 3A + 4| 5 = r = 3

The arc length and area of a sector are both 5, and the arc number of the central angle of the sector is 5______ ．