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For \mathrm{t \in(0,2 \pi)}, if \mathrm{ABC} is an equilateral triangle with vertices \mathrm{\mathrm{A}(\sin t,-\cos t), \mathrm{B}(\cos t, \sin \! t)} and \mathrm{C(a, b)} such that its orthocentre lies on a circle with centre \mathrm{\left(1, \frac{1}{3}\right),}then \mathrm{\left(a^{2}-b^{2}\right)} is equal to:

Option: 1

\frac{8}{3}


Option: 2

8


Option: 3

\frac{77}{9}


Option: 4

\frac{80}{9}


Answers (1)

best_answer

\mathrm{s \equiv \sin t, c \equiv \cos t}
Let orthocentre be \mathrm{(h, k)}
since it if an equilateral triangle hence orthocentre cor coincides with centroid
\mathrm{\therefore a+s+c=3 h\;, \;b+s-c=3 k}

\therefore(3 h-a)^2+(3 k-b)^2=(s+c)^2+(s-c)^2=2\left(s^2+c^2\right)=2 \\\therefore\left(k-\frac{a}{3}\right)^2+\left(k-\frac{b}{3}\right)^2=\frac{2}{4}

\\circle\: center \: at \left(\frac{a}{3}, \frac{b}{3}\right)\\ gives, \frac{a}{3}=1, \frac{b}{3}=\frac{1}{3} \Rightarrow a=3, b=1\\ \therefore a^2+a^2-b^2=8

Answer - B

Posted by

himanshu.meshram

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