When we get to the area formula of a circle When we deduce the area formula of a circle, we divide the circle into several equal parts to form an approximate rectangle. It is known that the area of a circle can be calculated when the length of the rectangle is 6.42 cm more than the width

When we get to the area formula of a circle When we deduce the area formula of a circle, we divide the circle into several equal parts to form an approximate rectangle. It is known that the area of a circle can be calculated when the length of the rectangle is 6.42 cm more than the width

The width of a rectangle is the radius of a circle. Let r = 3.14
The length of a rectangle is half the circumference of a circle
3.12×R=R+6.42
R=6.42/（3.14-1）=3
Circle area = 5 × 3 ^ 2 = 9

Formula of chord length

D -- the distance from the center of the circle to the chord, that is, the distance from the center of the chord
L - chord length
R -- radius of circle
R²=(l/2)²+d²
l²=4R²-4d²
l=2√(R²-d²)

What is the specific chord length formula?

If y ^ 2 = 2px, passing through the focus line intersection parabola at two points a (x1, Y1) and B (X2, Y2), then the chord length of AB = X1 + x2 + PD = √ (1 + k) | x1-x2 | = √ (1 + k) [(x1 + x2) 2 - 4x1x2] = √ (1 + 1 / k) | y1-y2 | = √ (1 + 1 / k) [(Y1 + Y2) 2 - 4y1y2] d = √ [(1 + k) △ / A;] = √ (1 + K;) (△) / | a |

Chord length formula of intersection of circle and line

Let the radius of the circle be r, the center of the circle be (m, n), and the linear equation be ax + by + C = 0
If the chord center distance is D, then d ^ 2 = (MA + Nb + C) ^ 2 / (a ^ 2 + B ^ 2)
Then the square of half the chord length is (R ^ 2-D ^ 2) / 2

The chord length formula of circle
Given that the radius of circle is r, the distance between chord centers is D, and the chord length is l, then L=

l=2√(r²-d²)

If the intersection of the line y = MX and the circle (x-3) ^ 2 + y ^ 2 = 4 is p, Q and the origin is O, then the value of | op | * | OQ | is?

Y = MX and (x-3) ^ 2 + y ^ 2 = 4
（m^2+1）x^2 - 6x+5=0;
Two are (x1, Y1) (X2, Y2)
x1x2=5/(m^2+1)
The required value is y1y2 + x1x2 = (m ^ 2 + 1) x1x2 = 5

If the intersection point of circle (x-3) ^ 2 + y ^ 2 = 4 and the line passing through the origin is known to be p, Q, then the trajectory equation of point P, q is

Let PQ: y = KX, P (x1, Y1) Q (X2, Y2) midpoint m (x0, Y0), and the circular equation, we get (1 + K ^ 2) x ^ 2-6x + 5 = 0,
X1 + x2 = 6 divided by (1 + K ^ 2) = 2x0. So K ^ 2 = (3 divided by x0) + 1 and m on PQ, Y0 ^ 2 = (kX0) ^ 2, substitute K ^ 2 = (3 divided by x0) + 1 to calculate Y0 ^ 2 + x0 ^ 2-3x0 = 0. Finally, it's done

Analytic geometry circle line
Circle x ^ 2 + y ^ 2 = 1 point a (- 2,0) point B (2, a) observe B from a so that the value of a is not blocked by the circle
According to the known tangent equation is y = radical 3 / 3 (x + 2), y = - radical 3 / 3 (x + 2), how do these two equations come from?
According to any formula,

The radius of a circle is 1, the distance from a point to o is 2, O and a form a right triangle, and a right side is half of OA, so when tangent, the inclination angle of the straight line is 30 ° or 150

Analytic geometry: line and circle
Let the equation of circle C x2 + y2 = R2 (r > 0), point m (x0, Y0) be a point in circle C, and o be the origin of coordinates, then the line x0x + y0y = R2
A. Away from circle C and perpendicular to line OM
B. Away from circle C and not perpendicular to line OM
C. Intersects circle C and is perpendicular to line OM
D. Intersecting circle C and not perpendicular to line OM
Please attach the detailed answer process

C in the hospital, the distance to the center of the circle is less than the radius, so √ (x0 & sup2; + Y0 & sup2;) 1 the distance from the center of the circle to the straight line = | 0 + 0-r & sup2;) / √ (x0 & sup2; + Y0 & sup2;) = R * r / √ (x0 & sup2; + Y0 & sup2;) > R * 1, that is, the distance from the center of the circle to the straight line is greater than the radius, so the skew ratio of the separated OM is Y0 / x0x0x + y0y = r2y =

The line L passing through the point m (1,2) and the circle C: (x-3) 2 + (y-4) 2 = 25 intersect at two points a and B, C is the center of the circle, when ∠ ACB is the minimum, the equation of the line L is ()
A. 2x+y-3=0B. x-y+1=0C. x+y-3=0D. 2x-y+3=0

Because point m (1,2) is in the interior of circle C: (x-3) 2 + (y-4) 2 = 25, it can be obtained from the intersection property of line AB and circle that when ∠ ACB is the smallest, the distance from center C to line AB is the largest, at this time, line AB is perpendicular to line MC. Because the slope of line MC is 4 − 23 − 1 = 1, then the slope of line L is - 1, and the equation of line L obtained from point oblique formula is Y-2 = - 1 (x-1), that is x + Y-3 = 0, So C