英语翻译如翻译不错立即采纳加分.其中很多公式不用翻译也不用管,只需翻英语部分.2.Proposed method2.1.

英语翻译
如翻译不错立即采纳加分.其中很多公式不用翻译也不用管,只需翻英语部分.
2.Proposed method
2.1.Chord properties
In this section,we introduce the property that relates the center of a circle to its chord and a point on the chord.
Fig.1(a) shows a circle and its chord AB.Let P be an interior point of the circle with its center at O and AB be the chord passing through P in a θ direction.The point A is defined as the endpoint in the quadrant I or II of the Cartesian coordinate system centered at P,and the point B defined as one in the quadrant III or IV.Let θ be a positive angle measured in a counterclockwise direction from PX to AB where θ ∈[0°,180°].Likewise,let φ be an angle measured from PX to OP.Then,
|AP|-|BP|=2|OP|cos(θ-φ).(1)
The proof is straightforward in the following.It is well-known that the perpendicular bisector of any chord of a circle passes through its center O (Rich,1963).Let C be the midpoint of the cord AB.
Then,
|AC|=|BC| and OC⊥AB.
So,
|CP|=|OP|cos(θ-φ).
|AP|-|BP|=(|AC|+|CP|)-(|BC|-|CP|)=2|CP|.
Therefore,
|AP|-|BP|=2|OP|cos(θ-φ)
In Fig.1(a),O(xo,yo) is related with the coordinates of P(xp; yp) as follows:
xo=xp+acosφ,
yo=yp-asinφ (2)
where a=|OP| and the minus sign in the equation of yo reflects that y value increases from top to bottom in the image coordinate system.If we have two chords passing through the point P,two equations are directly acquired from Eq.(1) so that a and φ can be solved with those two equations.As a result,(xo,yo) can be computed from Eq.(2).The details are given in the following section.
2.2.Computation for circle center
Consider two chords A1B1 and A2B2 passing through P(xp,yp) in the θ1 and θ2 directions (θ1≠θ2),respectively,as shown in Fig.1(b).
Let d1=|A1P|-|B1P| and d2=|A2P|-|B2P|.
Then,d1=2acos(θ1-φ) and d2=2acos(θ2-φ) from Eq.(1).
Solving for φ by dividing two sides of the equation of d1 by ones of the equation of d2 and expanding the cosine functions,we get
.一堆公式(略)
Eq.(4) shows us how the center of the circle can be computed from two intersecting chords.Specifically,if we have an interior point P of the circle and two chords passing through it,d1,d2,θ1 and θ2 given by these chords can locate the center of the circle from Eq.(4).One of the simplest cases is for θ1 = 0° and θ2 = 90°.
In that case,it can be easily shown that xo is the midpoint of the horizontal chord of θ1=0° and yo is the midpoint of the vertical chord of θ2=90°.This case is computationally efficient,but may not be robust because it uses only the vertical and horizontal chords.From this reason,the chords with randomly generated directions will be used along with Eq.(4) in this paper.
备注整理：至11-2-22,3 answers,1楼CV党,2楼朋友热心提议,3楼机器.