Question

The equation of the normal to the circle $x^2+y^2-8x-2y+12=0$ at the points whose ordinate is $-1$, will be

• A. $2x-y-7=0,2x+y-9=0$
• B. $2x+y-7=0,2x+y+9=0$
• C. $2x+y+7=0,2x+y+9=0$
• D. $2x-y+7=0,2x-y+9=0$

Answer 1

The correct option is A $2x-y-7=0,2x+y-9=0$

Circle $e{q}^n\to x^2+y^2-8x-2y+12=0....\left(i\right)$

We have draw normal whose ordinate is $=-1$

So for

So here $y=-1$

Putting $y=-1$ in $e{q}^n\left(i\right)$

$x^2+1-8x+2+12=0$

$x^2-8x+15=0$

$x^2-5x-3x+15=0\Rightarrow x(x-5)-3(x-5)=0$

$\Rightarrow (x-3)(x-5)=0$

$x=3,5$

So points where Normal to be drawn is

$(3,-1)$ and $(5,-1)$

For point $(3,-1)$, slope of Normal $=\frac{-1}{{\left(\frac{dy}{dx}\right)}_{(3,-1)}}$

from (i) $\frac{dy}{dx}$ is

differentiating $e{q}^n\left(i\right)$ with respect to $x$

$2x+2y\frac{dy}{dx}-8-2\frac{dy}{dx}+0=0.....\left(2\right)$

at point $(3,-1)$, put in $\left(2\right)$

$6-2\frac{dy}{dx}-8-2\frac{dy}{d}=0$

So slope of Normal $=\frac{-1}{\frac{-1}{2}}=2$

$e{q}^n$ will be $\Rightarrow (y+1)=2(x-3)$

$\Rightarrow y+1=2x-6$

$\Rightarrow 2x-y-7=0$

For point $(5,-1)$ slope of Normal from $e{q}^n\left(2\right)$

$10-2\frac{dy}{dx}-8-2\frac{dy}{dx}=0$

$2=4\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{1}{2}$

Slope of Normal $=\frac{-1}{{\left(\frac{dy}{dx}\right)}_{(5,-1)}}$

$=\frac{-1}{\left(\frac{1}{2}\right)}=-2$

So $e{q}^n$ of normal

$\Rightarrow (y+1)=-2(x-5)$

$\Rightarrow y+1=-2x+10$

$\Rightarrow 2x+y-9=0$

So $e{q}^n$ of Normals are $\Rightarrow (2x-y-7)=0$

and $2x+y-9=0$