Question

$A=(1,0),B(0,1)$ and $C$ is a point on the circle $x^2+y^2=1$. Then the locus of orthocentre of the triangle $ABC$ is

• A. $x^2+y^2=4$
• B. $x^2+y^2-x-y=4$
• C. $x^2+y^2-2x-2y+1=0$
• D. $x^2+y^2+2x-2y+1=0$

Answer 1

The correct option is D $x^2+y^2+2x-2y+1=0$

Let $C=(\cos \theta ,\sin \theta );A=(1,0);B=(0,1)$

Here the circumcenter $0$ is the origin $(0,0)$

Let the orthocenter be $H=(h,k)$

Here centroid $G$ is

$\Rightarrow G=(\frac{1+\cos \theta }{3},\frac{1+\sin \theta }{3})$

In a triangle we know that

$\Rightarrow 3G=2O+H$

Thus, $\Rightarrow G=(\frac{2O+h}{3},\frac{2O+k}{3})$

Thus, $\Rightarrow G=(\frac{h}{3},\frac{k}{3})$

$\Rightarrow (\frac{h}{3},\frac{k}{3})=(\frac{1+\cos \theta }{3},\frac{1+\sin \theta }{3})$

On comparing we get

$\Rightarrow h=1+\cos \theta \Rightarrow \cos \theta =h-1$

$\Rightarrow k=1+\sin \theta \Rightarrow \sin \theta =k-1$

We know that ${\sin }^2\theta +{\cos }^2\theta =1$

Thus $\Rightarrow (h-1{)}^2+(k-1{)}^2=1$

$\Rightarrow h^2-2h+1+k^2-2k+1=1$

$\Rightarrow h^2+k^2-2h-2k+1=0$

Hence the required locus can be obtained by replacing $(h,k)$ with $(x,y)$

$\Rightarrow x^2+y^2-2x-2y+1=0$