P.o.t.M. #1-Nov 2018 solution

IB Maths HL & SL / Analysis: P.o.t.M. #1-Nov 2018 solution

(a) The diagram at right (not to scale) shows the unique circle passing through \(\left( {1,0} \right)\), \(\left( {0,18} \right)\) and \(\left( {0,0} \right)\). These points are the vertices of a right triangle (shaded yellow). The hypotenuse of this right triangle must be the diameter of the circle and its midpoint \(\left( {\frac{1}{2},9} \right)\) is the centre of the circle. The diameter d of the circle can be computed using Pythagoras: \({d^2} = {1^2} + {18^2}\;\;\;\; \Rightarrow \;\;\;\;d = \sqrt {325} \)(b) The diagram below (to scale) shows...


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