Question

If the vertices of a variable triangle are $(3,4),$ $(5\cos \theta ,5\sin \theta )$ and $(5\sin \theta ,-5\cos \theta ),$ then the 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 B $(x+y-7{)}^2+(x-y+1{)}^2=100$

Given coordinates of the triangle are
$A(3,4),B(5\cos \theta ,5\sin \theta )$ and $C(5\sin \theta ,-5\cos \theta )$

Let the circumcentre be $S=(x,y)$.
Then $S{A}^2=S{B}^2=S{C}^2$
Taking $S{B}^2=S{C}^2$, we get
$(x-5\cos \theta {)}^2+(y-5\sin \theta {)}^2=(x-5\sin \theta {)}^2+(y+5\cos \theta {)}^2\Rightarrow -x\cos \theta -y\sin \theta =-x\sin \theta +y\cos \theta \Rightarrow x(\sin \theta -\cos \theta )=y(\sin \theta +\cos \theta )\cdots \left(1\right)$

Taking $S{A}^2=S{B}^2$, we get
$(x-3{)}^2+(y-4{)}^2=(x-5\cos \theta {)}^2+(y-5\sin \theta {)}^2\Rightarrow -6x-8y=-10x\cos \theta -10y\sin \theta \Rightarrow (10\cos \theta -6)x=(8-10\sin \theta )y\cdots \left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we get
$x=0,y=0\Rightarrow S=(0,0)$

Now, the centroid $G=(\frac{3+5\cos \theta +5\sin \theta }{3},\frac{4+5\sin \theta -5\cos \theta }{3})$
Assume the orthocentre to be $H=(h,k)$.
$\because$ Centroid $\left(G\right)$ divides the line joining orthocentre $\left(H\right)$ and circumcentre $\left(O\right)$ in ratio $2:1$,
$G=(\frac{0+h}{3},\frac{0+k}{3})\Rightarrow (\frac{3+5\cos \theta +5\sin \theta }{3},\frac{4+5\sin \theta -5\cos \theta }{3})=(\frac{h}{3},\frac{k}{3})\Rightarrow h=3+5\cos \theta +5\sin \theta ,k=4+5\sin \theta -5\cos \theta$

Now, $h+k=7+10\sin \theta$
$h-k=-1+10\cos \theta$
Therefore, $(h+k-7{)}^2+(h-k+1{)}^2=(10\sin \theta {)}^2+(10\cos \theta {)}^2$

Hence, the locus of the orthocentre is $(x+y-7{)}^2+(x-y+1{)}^2=100$