Question

Let $AB$ be the chord of contact of the point (5,-5) w.r.t. the circle $x^2+y^2=5,$ then find the locus of the orthocentre of the triangle $PAB,$ where $P$ be any point moving on the circle.

Answer 1

Equation of chord of contact $AB$ is $5x-5y=5$

Solving it with $x^2+y^2=5$ we get,

$x^2+(x-1{)}^2=5$

$\Rightarrow x^2-x-2=0\Rightarrow (x-2)(x+1)=0$

$\Rightarrow x=-1,2\text{and}y=-2,1$

So, $A$ and $B$ are (-1,-2) and (2,1)Let

$P\equiv (\sqrt{5}\cos \theta ,\sqrt{5}\sin \theta )$

Now as circumcentre of the triangle $PAB$ is origin, orthocentre would be $h=x_1+x_2+x_3$ and $k=y_1+y_2+y_3$ (as centroid divides line joining orthocentre and circumcentre in 2: 1 ).i.e.

$h=\sqrt{5}\cos \theta +1$ and

$k=\sqrt{5}\sin \theta -1$

So the required locus is $(x-1{)}^2+(y+1{)}^2=5$