question_answer

Find the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

Answer:

Let a be the radius of family of circles in the first quadrant, which touch the coordinate axes.                         \[\therefore \] Coordinates of centre of circle = (a, a) We know that equation of circle which has centre (h, k) and radius r is given by             \[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\]                 ?(i) Here, we have (h, k) = (a, a) and r = a \[\therefore \] Equation of family of such circles is             \[{{(x-a)}^{2}}+{{(y-a)}^{2}}={{a}^{2}}\]             ?(ii) On differentiating both sides w.r.t. x, we get             \[2(x-a)+2(y-a)\frac{dy}{dx}=0\] \[\Rightarrow \]   \[x-a+(y-a)\cdot y'=0\]    \[\left[ \because \,\,\,\frac{dy}{dx}=y' \right]\] \[\Rightarrow \]   \[x+yy'=a+ay'\] \[\therefore \]      \[a=\frac{x+yy'}{1+y'}\] On putting above value of a in Eq. (ii), we get \[{{\left( x-\frac{x+yy'}{y'+1} \right)}^{2}}+{{\left( y-\frac{x+yy'}{y'+1} \right)}^{2}}={{\left( \frac{x+yy'}{y'+1} \right)}^{2}}\] \[\Rightarrow \] \[{{\left[ \frac{xy'+x-x-yy'}{y'+1} \right]}^{2}}+{{\left[ \frac{yy'+y-x-yy'}{y'+1} \right]}^{2}}\]                                                 \[={{\left( \frac{x+yy'}{y'+1} \right)}^{2}}\] On multiplying both sides by \[{{(y'+1)}^{2}}\] we get             \[{{(xy'-yy')}^{2}}+{{(y-x)}^{2}}={{(x+yy')}^{2}}\] \[\Rightarrow \]   \[{{(x-y)}^{2}}{{(y')}^{2}}+{{(x-y)}^{2}}={{(x+yy')}^{2}}\] \[[\because \,\,\,{{(x-y)}^{2}}={{(y-x)}^{2}}]\] \[\therefore \]      \[{{(x-y)}^{2}}[{{(y')}^{2}}+1]={{(x+yy')}^{2}}\] Which is the required differential equation.