A) \[{{\left[ 1+{{\left( \frac{dy}{dx} \right)}^{3}} \right]}^{2}}={{r}^{2}}{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{2}}\]
B) \[{{\left[ 1+{{\left( \frac{dy}{dx} \right)}^{3}} \right]}^{2}}={{r}^{2}}{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{3}}\]
C) \[{{\left[ 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right]}^{3}}={{r}^{2}}{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{2}}\]
D) \[{{\left[ 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right]}^{3}}={{r}^{2}}{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{3}}\]
Correct Answer: C
Solution :
The equation of the family of circles of radius \[r\]is \[{{(x-a)}^{2}}+{{(y-b)}^{2}}={{r}^{2}}\] ?(i) where a and b are arbitrary constants. On differentiating Eq. (i) w.r.t. \[x,\]we get \[2(x-a)+2(y-b)\frac{dy}{dx}=0\] \[\Rightarrow \] \[(x-a)+(y-b)\frac{dy}{dx}=0\] ?(ii) On differentiating Eq. (ii) w.r.t. \[x,\]we get \[1+(y-b)\frac{{{d}^{2}}y}{d{{x}^{2}}}+{{\left( \frac{dy}{dx} \right)}^{2}}=0\] \[\Rightarrow \] \[(y-b)=-\frac{1+{{(dy/dx)}^{2}}}{{{d}^{2}}y/d{{x}^{2}}}\] ?(iii) On putting the value of \[(y-b)\] in Eq. (ii), we get \[(x-a)=\frac{\left[ 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right]\frac{dy}{dx}}{\frac{{{d}^{2}}y}{d{{x}^{2}}}}\] ?(iv) On putting the value of\[(y-b)\]and \[(x-a),\] Eq. (i), we get \[\frac{{{\left[ 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right]}^{2}}{{\left( \frac{dy}{dx} \right)}^{2}}}{{{({{d}^{2}}y/d{{x}^{2}})}^{2}}}+\frac{\left[ 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right]}{{{({{d}^{2}}y/d{{x}^{2}})}^{2}}}={{r}^{2}}\] \[\Rightarrow \] \[{{\left[ 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right]}^{3}}={{r}^{2}}\left[ \frac{{{d}^{2}}{{y}^{2}}}{d{{x}^{2}}} \right]\] This is the required differential equation.
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