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.
$\therefore a+s+c=3h,b+s-c=3k$
$\therefore (3h-a{)}^2+(3k-b{)}^2=(s+c{)}^2+(s-c{)}^2=2\left(s^2+c^2\right)=2$
$\therefore {\left(h-\frac{a}{3}\right)}^2+{\left(K-\frac{b}{3}\right)}^2=\frac{2}{9}$
$\text{ circle centre at }\left(\frac{a}{3},\frac{b}{3}\right)$
$\text{ Gives, }\frac{a}{3}=1,\frac{b}{3}=\frac{1}{3}\Rightarrow a=3,b=1$
$\therefore a^2-b^2=8$