Question

The normal at each point of a curve passes through $(3,1)$. If the point $(1,1)$ lies on the curve the equation of the curve is:

• A. $x^2+y^2=2$
• B. $2x^2-y^2=1$
• C. $(x-3{)}^2+(y-1{)}^2=4$
• D. $(x-1{)}^2+(y-3{)}^2=4$

Answer 1

The correct option is C $(x-3{)}^2+(y-1{)}^2=4$

The equation of normal at $(x,y)$ to the curve having slope $\frac{dy}{dx}$ is $y-y=\frac{1}{-\frac{dy}{dx}}\times (x-x)$
But the normal passes through $(3,1)$.
$\Rightarrow 1-y=\frac{1}{-\frac{dy}{dx}}\times (3-x)$

$\Rightarrow \frac{dy}{dx}(1-y)=(x-3)$$\Rightarrow (1-y)dy=(x-3)dx$

$\Rightarrow (x-3)dx+(y-1)dy=0$

On integrating both sides, we get

$\Rightarrow \frac{(y-1{)}^2}{2}+\frac{(x-3{)}^2}{2}=C$

But the given curve passes through $(1,1)$.

On substituting, $C=4$, the required equation of curve is $\frac{(x-3{)}^2}{2}+\frac{(y-1{)}^2}{2}=4$