#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise  Revision Exercise Question 7 Maths Textbook Solution.

$\left ( x^{2}-9 \right )\left ( {y}' \right )^{2}+x^{2}=0$

Hint:

You must know about the equation of circle

Given:

Family of circles having centre on y-axis and radius 3 units

Solution:

General equation of circle is

$\left ( x-a \right )^{2}+\left ( y-b \right )^{2}=r^{2}$

Given centre is on y-axis

$\therefore$           Centre$= \left ( 0,b \right )$  and Radius$= 3$

Hence, our equation is

$\left ( x-0 \right )^{2}+\left ( y-b \right )^{2}=3^{2}$

$x^{2}+\left ( y-b \right )^{2}=9$

Differentiate with respect to $x$ ,

\begin{aligned} &2 x+2(y-b) y^{\prime}=0 \\ &2\left(x+(y-b) y^{\prime}\right)=0 \\ &(y-b) y^{\prime}=-x \\ &(y-b)=\frac{-x}{y^{\prime}} \end{aligned}

Put value in (i)

\begin{aligned} &x^{2}+(y-b)^{2}=9 \\ &x^{2}+\left[\frac{-x}{y^{\prime}}\right]^{2}=9 \\ &x^{2}+\frac{x^{2}}{y^{\prime 2}}=9 \\ &\frac{x^{2}\left(y^{\prime}\right)^{2}+x^{2}}{\left(y^{\prime}\right)^{2}}=9 \\ &x^{2}\left(y^{\prime}\right)^{2}+x^{2}=9\left(y^{\prime}\right)^{2} \\ &x^{2}\left(y^{\prime}\right)^{2}-9\left(y^{\prime}\right)^{2}+x^{2}=0 \\ &\left(y^{\prime}\right)^{2}\left[x^{2}-9\right]+x^{2}=0 \\ &\therefore \left(x^{2}-9\right)\left(y^{\prime}\right)^{2}+x^{2}=0 \end{aligned}