Question
Vertices of a variable triangle are $$\left( {3,\,4} \right),\,\left( {5\,\cos \,\theta ,\,5\,\sin \,\theta } \right)$$ and $$\left( {5\,\sin \,\theta ,\, - 5\,\cos \,\theta } \right),$$ where $$\theta \, \in \,R.$$ Locus of it's orthocentre is :
A.
$${\left( {x + y - 1} \right)^2} + {\left( {x - y - 7} \right)^{^2}} = 100$$
B.
$${\left( {x + y - 7} \right)^2} + {\left( {x - y - 1} \right)^{^2}} = 100$$
C.
$${\left( {x + y - 7} \right)^2} + {\left( {x + y - 1} \right)^{^2}} = 100$$
D.
$${\left( {x + y - 7} \right)^2} + {\left( {x - y + 1} \right)^{^2}} = 100$$
Answer :
$${\left( {x + y - 7} \right)^2} + {\left( {x - y + 1} \right)^{^2}} = 100$$
Solution :
Distance of all the points from $$\left( {0,\,0} \right)$$ are $$5$$ unit. That means circumcentre of the triangle formed by the given point is $$\left( {0,\,0} \right).$$ If $$G\left( {h,\,k} \right)$$ be the centroid of triangle, then
$$3h = 3 + 5\left( {\cos \,\theta + \sin \,\theta } \right),\,3k = 4 + 5\left( {\sin \,\theta - \cos \,\theta } \right)$$
If $$H\left( {\alpha ,\,\beta } \right)$$ be the orthocentre, then
$$\eqalign{
& OG:GH = 1:2\, \Rightarrow \alpha = 3h,\,\,\beta = 3k \cr
& \cos \,\theta + \sin \,\theta = \frac{{\alpha - 3}}{5},\,\sin \theta - \cos \,\theta = \frac{{\beta - 4}}{5} \cr
& \Rightarrow \sin \,\theta = \frac{{\alpha + \beta - 7}}{{10}},\,\,\cos \,\theta = \frac{{\alpha - \beta + 1}}{{10}} \cr} $$
Thus, locus of $$\left( {\alpha ,\,\beta } \right)$$ is $${\left( {x + y - 7} \right)^2} + {\left( {x - y + 1} \right)^{^2}} = 100.$$