Question

Vertices of a variable triangle are $(3,4),(5\cos \theta ,5\sin \theta )$ and $(5\sin \theta ,-5\cos \theta )$, where $\theta \in R$. Locus of its orthocentre is

• A. $(x+y-1{)}^2+(x-y-7{)}^2=100$
• B. $(x+y-7{)}^2+(x-y-1{)}^2=100$
• C. $(x+y-7{)}^2+(x+y-1{)}^2=100$
• D. $(x+y-7{)}^2+(x-y+1{)}^2=100$

Answer 1

The correct option is D $(x+y-7{)}^2+(x-y+1{)}^2=100$

Distance of all the vertices from $(0,0)$ is $5$ units.

That means the circumcentre of the triangle formed by the given points is $(0,0)$.

If $H$ be the orthocentre, then

$OG:GH=1:2$

If $G(h,k)$ be the centroid of the triangle, then

$G(\frac{3+5\cos \theta +5\sin \theta }{3},\frac{4+5\sin \theta -5\cos \theta }{3})=G(\frac{0+h}{3},\frac{0+k}{3})$

$\therefore \frac{(h-3)}{5}=(\cos \theta +\sin \theta ),\frac{k-4}{5}=(\sin \theta -\cos \theta )$.

$\Rightarrow \cos \theta +\sin \theta =\frac{\alpha -3}{5}$, $\sin \theta -\cos \theta =\frac{\beta -4}{5}$

$\Rightarrow \sin \theta =\frac{\alpha +\beta -7}{10}$

$\Rightarrow \cos \theta =\frac{\alpha -\beta +1}{10}$

Squaring and adding, we get
$(x+y-7{)}^2+(x-y+1{)}^2=100$ ....as ${\sin }^2\theta +{\cos }^2\theta =1$