question_answer

Form the differential equation of the family of circles touching the X-axis at origin.

Answer:

Let C denotes the family of circles touching X-axis at origin. Again, let (0, a) be the coordinates of the centre of any member of the family.              Then, the equation of family C is             \[{{x}^{2}}+{{(y-a)}^{2}}={{a}^{2}}\] \[\Rightarrow \]   \[{{x}^{2}}+{{y}^{2}}+{{a}^{2}}-2ay={{a}^{2}}\] \[\Rightarrow \]   \[{{x}^{2}}+{{y}^{2}}=2ay\]                                       ?(i) Where a is arbitrary constant. Now. Differentiating both sides w.r.t. x, we get \[2x+2y\frac{dy}{dx}=2a\frac{dy}{dx}\] \[\Rightarrow \] \[x+y\frac{dy}{dx}=a\frac{dy}{dx}\]      \[\Rightarrow \]   \[a=\frac{x+y{{y}_{1}}}{{{y}_{1}}}\]             \[\left[ \because \,\,\,\frac{dy}{dx}={{y}_{1}} \right]\] Now, substituting the value of a in Eq. (i), we get             \[{{x}^{2}}+{{y}^{2}}=2y\left[ \frac{x+y{{y}_{1}}}{{{y}_{1}}} \right]\] \[\Rightarrow \]   \[{{y}_{1}}({{x}^{2}}+{{y}^{2}})=2xy+2{{y}^{2}}{{y}_{1}}\] \[\Rightarrow \]   \[{{y}_{1}}({{x}^{2}}+{{y}^{2}}+2{{y}^{2}})=2xy\] \[\Rightarrow \]   \[{{y}_{1}}=\frac{2xy}{{{x}^{2}}-{{y}^{2}}},\] Which is the required differential equation of the given family of circles.