Question

The equation of the normals to the circle $x^2+y^2-8x-2y+12=0$ at the points whose ordinate is $-1$ is/are

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

Answer 1

Answer: (A), (B)

$2x-y-7=0$

Given circle is $x^2+y^2-8x-2y+12=0$
Putting the ordinate, we get the abscissa
$x^2-8x+15=0\Rightarrow (x-5)(x-3)=0\Rightarrow x=3,5$
$\therefore$ Points are $(5,-1)$ and $(3,-1)$

Centre of the circle is $(4,1)$
Normal is given by
$\frac{x-5}{5-4}=\frac{y+1}{-1-1}\Rightarrow 2x+y-9=0$
And
$\frac{x-3}{3-4}=\frac{y+1}{-1-1}\Rightarrow 2x-y-7=0$