Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.
Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.
The equation of a circle in the first quadrant with centre (a, a) and radius (a) which touches the coordinate axes is:
$(x-a)^{2}+(y-a)^{2}=a^{2}$ ...(1)
Differentiating equation (1) with respect to x, we get:
$2(x-a)+2(y-a) \frac{d y}{d x}=0$
$\Rightarrow(x-a)+(y-a) y^{\prime}=0$
$\Rightarrow x-a+y y^{\prime}-a y^{\prime}=0$
$\Rightarrow x+y y^{\prime}-a\left(1+y^{\prime}\right)=0$
$\Rightarrow a=\frac{x+y y^{\prime}}{1+y^{\prime}}$
Substituting the value of a in equation (1), we get:
$\left[x-\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right)\right]^{2}+\left[y-\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right)\right]^{2}=\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right)^{2}$
$\Rightarrow\left[\frac{(x-y) y^{\prime}}{\left(1+y^{\prime}\right)}\right]^{2}+\left[\frac{y-x}{1+y^{\prime}}\right]^{2}=\left[\frac{x+y y^{\prime}}{1+y^{\prime}}\right]^{2}$
$\Rightarrow(x-y)^{2} \cdot y^{\prime 2}+(x-y)^{2}=\left(x+y y^{\prime}\right)^{2}$
$\Rightarrow(x-y)^{2}\left[1+\left(y^{\prime}\right)^{2}\right]=\left(x+y y^{\prime}\right)^{2}$
Hence, the required differential equation of the family of circles is $(x-y)^{2}\left[1+\left(y^{\prime}\right)^{2}\right]=\left(x+y y^{\prime}\right)^{2} .$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.