maths~!!!

(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