Question

For $t \in(0,2 \pi)$, if $ABC$ is an equilateral triangle with vertices $A ( \sin t, -{\cos t}), B ( \cos t, \sin t )$ 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 :

• A. $\frac{8}{3}$
• B. 8
• C. $\frac{77}{9}$
• D. $\frac{80}{9}$

Answer 1

Answer: (B)

$s \equiv \sin t , c \equiv \cos t$
Let orthocentre be $( h , k )$
Since it if an equilateral triangle hence orthocentre coincides with centroid.
$\therefore a + s + c =3 h , b + s - c =3 k$
$ \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( h -\frac{ a }{3}\right)^2+\left( K -\frac{ b }{3}\right)^2=\frac{2}{9} $
$ \text { circle centre at }\left(\frac{ a }{3}, \frac{ b }{3}\right)$
$ \text { Gives, } \frac{ a }{3}=1, \frac{ b }{3}=\frac{1}{3} \Rightarrow a =3, b =1 $
$ \therefore a ^2- b ^2=8$