question_answer

If the vertices of a triangle are (12,5), \[(13\,\sin \,\theta ,\,-13\cos \theta )\] and \[\left( -13\cos \theta ,\,13\sin \theta  \right),\] where \[\theta \] is arbitrary, then locus of its orthocentre is

A)  \[x-y=7\]                           

B)  \[x+y=17\]

C)  \[{{\left( x-12 \right)}^{2}}+{{\left( y-5 \right)}^{2}}={{13}^{2}}\]

D)  \[{{\left( x-12 \right)}^{2}}+{{\left( y-5 \right)}^{2}}=2\times {{13}^{2}}\]

Correct Answer: A

Solution :

All three vertices lie on \[{{x}^{2}}+{{y}^{2}}=169={{13}^{2}}\] Hence, circumcircle of \[\Delta \] is \[O(0,\,0)\]. Centriod is  \[G\left( \frac{12+13\,\sin \theta -13\cos \theta }{3},\frac{5-13\,\sin \theta +13\cos \theta }{3} \right)\]   \[\Rightarrow \] H divides OG in 3 : 2 externally. \[H(12+13\,\sin \theta -13\,\cos \theta ,\,5-13\cos \theta +13\sin \theta )\] Whose locus is \[x-y=7\].