A) \[{{\left( y-\frac{dy}{dx} \right)}^{2}}={{a}^{2}}\left[ 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right]\]
B) \[{{\left( y-x\frac{dy}{dx} \right)}^{2}}={{a}^{2}}\left[ 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right]\]
C) \[\left( y-x\frac{dy}{dx} \right)={{a}^{2}}\left[ 1+\frac{dy}{dx} \right]\]
D) \[\left( y-\frac{dy}{dx} \right)={{a}^{2}}\left[ 1-\frac{dy}{dx} \right]\]
Correct Answer: B
Solution :
The equation of straight line touching the circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]is \[x\cos \theta +y\sin \theta =a\] ?(i) On differentiating w.r.t. x, regarding\[\theta \]as a constant \[\cos \theta +y\sin \theta =0\] ?(ii) From Eqs. (i) and (ii), we get \[\cos \theta =\frac{ay}{xy-y}\]and \[\sin \theta =-\frac{a}{xy-y}\] \[\because {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\] \[\therefore \] \[\frac{{{a}^{2}}y{{}^{2}}+{{a}^{2}}}{{{(xy-y)}^{2}}}=1\] \[\Rightarrow \] \[{{\left( y-x\frac{dy}{dx} \right)}^{2}}={{a}^{2}}\left[ 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right]\]
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