Question

The radius of the circle $ {{(x\,\cos \theta +y\sin \theta -a)}^{2}}+(x\,\sin \theta -y\cos \theta )={{k}^{2}} $ is

• A. $ {{a}^{2}}+{{b}^{2}}-{{k}^{2}} $
• B. $ a\,sin\,\theta -b\,\cos \,\theta $
• C. $ {{a}^{2}}+{{b}^{2}} $
• D. $ k $

Answer 1

Answer: (D)

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 $