Question

The order and degree of the differential equation of all circles in the first quadratic which touch the co-ordinate axis is:

• A. $1,2$
• B. $2,1$
• C. $3,2$
• D. $4,3$

Answer 1

Answer: (A)

$1,2$

Equation will be

$(x-p{)}^2+(y-p{)}^2=p^2$

Now, differentiate with respect to x,

$2(x-p)+2(y-p)\frac{dy}{dx}=0$

$\Rightarrow p=\frac{x+y\frac{dy}{dx}}{1+\frac{dy}{dx}}=\frac{xdx+ydy}{dx+dy}$

$\Rightarrow (x-p{)}^2+(y-p{)}^2=p^2$

$\Rightarrow (xdy-ydy{)}^2+(ydx-xdx{)}^2=(xdx+ydy{)}^2$

Now, on further solving we get

$(xdy{)}^2+(ydx{)}^2=6xydxdy$

Now, dividing by (ydx)^{2}, we get

$(\frac{x}{y}\frac{dy}{dx}{)}^2+1=6\frac{x}{y}\frac{dy}{dx}$

Hence, Degree=$2$, Order=$1$