A) \[{{\alpha }^{2}}+{{\beta }^{2}}\]
B) \[{{\alpha }^{2}}+{{\beta }^{2}}-{{\alpha }^{2}}\]
C) \[{{\alpha }^{2}}\]
D) \[{{\alpha }^{2}}+{{\beta }^{2}}+{{\alpha }^{2}}\]
Correct Answer: B
Solution :
[b] Any point on the line at a distance at a r from the point \[P(\alpha ,\beta )\] is \[(\alpha +rcos\theta ,\beta +r\,sin\theta )\] If this point lies on \[{{x}^{2}}+{{y}^{2}}={{a}^{2}},\] then \[{{\alpha }^{2}}+{{r}^{2}}{{\cos }^{2}}\theta +2ar\cos \theta +{{\beta }^{2}}+{{r}^{2}}{{\sin }^{2}}\theta +2\beta r\sin \theta ={{a}^{2}}\]\[\Rightarrow {{r}^{2}}+2r(\alpha cos\theta +\beta sin\theta )+{{\alpha }^{2}}+{{\beta }^{2}}={{a}^{2}}\] \[\Rightarrow {{r}^{2}}+2r(\alpha cos\theta +\beta sin\theta )+{{\alpha }^{2}}+{{\beta }^{2}}-{{a}^{2}}=0\] Nor, if \[PA={{r}_{1}}\] and \[PB={{r}_{2}},\] then \[{{r}_{1}}\] and \[{{r}_{2}}\] must be roots of this equation. \[\therefore \,\,\,\,\,PA.PB={{r}_{1}}.{{r}_{2}}={{\alpha }^{2}}+{{\beta }^{2}}-{{a}^{2}}\]
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