Question

The differential equation by eliminating $a,b$ from $(x-a{)}^2+(y-b{)}^2=r^2$ is

• A. $(1-(y_1{)}^2{)}^3=r^2(y_2{)}^2$
• B. $(1+(y_1{)}^2{)}^3=r^2(y_2{)}^2$
• C. $(1+(y_1{)}^2{)}^3=r{y}^2$
• D. $y_1^2=r{y}^2$

Answer 1

Answer: (B)

$(1+(y_1{)}^2{)}^3=r^2(y_2{)}^2$

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

By differentiating this equation, we get

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

$a=x+(y-b)\frac{dy}{dx}$-Equation 1

Differentiating again, we get

$0=1+(y-b)\frac{d^{2}y}{d{x}^2}+(\frac{dy}{dx}{)}^2$

$-b=\frac{(-1-(\frac{dy}{dx}{)}^2)}{\frac{d^{2}y}{d{x}^2}}-y=\frac{-1-y_1^2}{y_2}-y$-Equation 2

Now, substituting the values of a & b,

in $(x-a{)}^2+(y-b{)}^2=r^2$, we get

$(x-x-(y-b)\frac{dy}{dx}{)}^2+(y-b{)}^2=r^2$

$(y-b{)}^2(\frac{dy}{dx}{)}^2+(y-b{)}^2=r^2$

$1+y_1^2=\frac{r^2}{(y-b{)}^2}$

and $y-b=\frac{-1-y_1^2}{y_2}$

$1+y_1^2=\frac{r^2{y}_2^2}{(1+y_1^2{)}^2}$

$(1+y_1^2)(1+y_1^2{)}^2=r^2{y}_2^2$

$(1+y_1^2{)}^3=r^2{y}_2^2$