A) \[x+y=0\]
B) \[x-y=0\]
C) \[xy=0\]
D) None of these
Correct Answer: C
Solution :
Let the coordinates of\[P\]be\[(h,\,\,k)\].Let the equation of a tangent from\[P(h,\,\,k)\]to the circle\[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]be \[y=mx+a\sqrt{1+{{m}^{2}}}\] Since\[P(h,\,\,k)\]lies on \[y=mx+a\sqrt{1+{{m}^{2}}}\] therefore, \[k=mh+a\sqrt{1+{{m}^{2}}}={{(k-mh)}^{2}}={{a}^{2}}(1+{{m}^{2}})\]this is a quadratic equation in m. Let the two roots be\[{{m}_{1}}\]and\[{{m}_{2}}\], then \[{{m}_{1}}+{{m}_{2}}=\frac{2hK}{{{K}^{2}}-{{a}^{2}}}\] But\[\tan \alpha ={{m}_{1}},\,\,\tan \beta ={{m}_{2}}\]and it is given that \[\cot \alpha +\cot \beta =0\] \[\therefore \] \[\frac{1}{{{m}_{1}}}+\frac{1}{{{m}_{2}}}=0\] \[\Rightarrow \] \[{{m}_{1}}+{{m}_{2}}=0\] \[\Rightarrow \] \[\frac{2hK}{{{K}^{2}}-{{a}^{2}}}=0\] \[\Rightarrow \] \[hK=0\] Hence, the locus of\[(h,\,\,K)\]is\[xy=0\].
You need to login to perform this action.
You will be redirected in
3 sec
