A) \[{{a}^{2}}+{{b}^{2}}-{{k}^{2}}\]
B) \[a\,sin\,\theta -b\,\cos \,\theta \]
C) \[{{a}^{2}}+{{b}^{2}}\]
D) \[k\]
Correct Answer: D
Solution :
Given equation of circle is \[{{(x\,\cos \,\theta +y\,\sin \theta -a)}^{2}}+{{(x\,\sin \theta -y\,\cos \theta -b)}^{2}}\] \[={{k}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}{{\cos }^{2}}\theta +{{y}^{2}}{{\sin }^{2}}\theta +2xy\,\sin \theta .\cos \theta +{{a}^{2}}\] \[-2a(x\,\cos \,\theta +y\,\sin \theta )+{{x}^{2}}{{\sin }^{2}}\theta +{{y}^{2}}\,{{\cos }^{2}}\theta \] \[-2xy\,\sin \theta .\cos \theta +{{b}^{2}}-2b\,(x\,\sin \theta -y\,cos\,\theta )={{k}^{2}}\]\[\Rightarrow \] \[{{x}^{2}}\,({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )+{{y}^{2}}({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )\] \[+(-2a\,\cos \theta -2b\sin \theta )x+(-2a\,\sin \theta +2b\,\cos \theta )y\] \[+({{a}^{2}}+{{b}^{2}}-{{k}^{2}})=0\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-2(a\,\cos \,\theta +b\sin \theta )x\] \[+2(-a\,\sin \theta +b\cos \theta )y+({{a}^{2}}+{{b}^{2}}-{{k}^{2}})=0\] ?..(i) \[(\because \,\,\,\,\,\,{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1)\] On comparing with \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0,\] we get \[g=-(a\,\cos \,\theta +b\,\sin \theta ),\] \[f=+(-a\,\sin \theta +b\,\cos \,\theta )\] and \[c={{a}^{2}}+{{b}^{2}}-{{k}^{2}}\] \[\therefore \] Radius \[=\sqrt{{{g}^{2}}+{{f}^{2}}-c}\] \[=\sqrt{\begin{align} & {{(a\,\cos \theta +b\,\sin \,\theta )}^{2}}+{{(-a\,\sin \theta +\,b\,\cos \theta )}^{2}} \\ & -({{a}^{2}}+{{b}^{2}}-{{k}^{2}}) \\ \end{align}}\] \[=\sqrt{{{a}^{2}}\,{{\cos }^{2}}\theta +{{b}^{2}}\,{{\sin }^{2}}\,\theta +2ab\,\sin \theta .\cos \,\theta }\] \[+{{a}^{2}}\,{{\sin }^{2}}\theta +{{b}^{2}}\,{{\cos }^{2}}\theta -2ab\,\sin \theta .\cos \theta \] \[-({{a}^{2}}+{{b}^{2}}-{{k}^{2}})\] \[\Rightarrow \] \[\sqrt{{{a}^{2}}+{{b}^{2}}-{{a}^{2}}-{{b}^{2}}+{{k}^{2}}}=\sqrt{{{k}^{2}}}=k\]
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