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maths~!!!
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f4-f5數學的圓形題目~! 圖片參考：http://g.imagehost.org/0736/ScreenHunter_04_Feb_27_10_44.gif
最佳解答:
(a i) Since the circles are equal, they have the same radius, i.e. SR = SP = QP = QR Join SQ and it will be equal to the 4 sides mentioned above too (radius of the circle) Therefore △RQS and △PQS are congurent equilateral triangles and so RS//QP and RQ//SP So PQRS is a rhombus. (ii) ∠OBP = ∠OAP (Isos. △OAB) ∠BQP = ∠OBP and ∠ASP = ∠OAP (∠s in alt. segment) So ∠BQP = ∠ASP and so BP = AP for the reason of equal angle at circumference, equal chord. Join OP, the with OB = OA given, △OBP and △OAP are congurent (SAS) So OP is perp. to AB Join BS and AQ, then △SPB and △QPA are congurent (SSS) Therefore ∠OPA = ∠SPB and also, when joining RP, it bisects ∠SPQ and therefore ∠BPR = 90 So RP is also perp. to BA and so OPR is a straight line. (b i) OB = 5, then A is (5, 0) since it lies on the x-axis. Thus P is the mid-point of AB which is (1, 2). (ii) From (a), we can see that △QPA is an equil. triangle and therefore the radius of the circle is in fact = AP which is √20 And since the circle touches the x-axis at A, Q, the centre, is directly above A with location at (0, √20) So equation of the circle is: (x - 0)2 + (y - √20)2 = 20 x2 + y2 - 2√20 y = 0 x2 + y2 - 4√5 y = 0
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