For \[\mathrm{t} \in(0,2 \pi)\], if \[\mathrm{ABC}\] is an equilateral triangle with vertices \[\mathrm{A}(\sin t,-\cos \mathrm{t}), \mathrm{B}(\operatorname{cost}, \sin t)\] and C(a,b) such that its orthocentre lies on a circle with centre \[\left(1, \frac{1}{3}\right)\], then \[\left(a^{2}-b^{2}\right)\] is equal to :
Solutions · 1
Official solution
Let P(h,k) be the orthocentre of \[\Delta\]ABC
Then
\[h = {{\sin t + \cos t + a} \over 3},\,k = {{ - \cos t + \sin t + b} \over 3}\]
(Orthocentre coincide with centroid)
\[\therefore\] (3h−a)2+(3k−b)2=2
\[\therefore\] \[{\left( {h - {a \over 3}} \right)^2} + {\left( {k - {b \over 3}} \right)^2} = {2 \over 9}\]
\[\because\] Orthocentre lies on circle with centre \[\left( {1,{1 \over 3}} \right)\]
\[\therefore\] \[a = 3,\,b = 1\]
\[\therefore\] a2−b2=8